MTH 254H
Honors Multivariable Calculus
Section 2 − Fall 2022

• Theorem: The curl of a vector field measures its rate of circulation around a point, per area enclosed.
  For a vector field
  F(x,y) = (p(x,y), q(x,y)),
  a chosen point (a,b), and lengths Δx, Δy, let c(t) be the rectangular curve going from (a,b) to (a+Δx, b) to (a+Δx, b+Δy) to (a, b+Δy) and back to (a,b). Then we have:
  lim_{Δx,Δy → 0}   ¹⁄_ΔxΔy   ∮ F(c) • dc   =   ^∂q⁄_∂x(a,b) − ^∂p⁄_∂y(a,b).

  Here the corner (a,b) stays constant while Δx and Δy independently approach zero.

• Proof. We compute the four segments of the line integral:

  ∮ F(c) • dc	=	∫0Δx F(a+t,b) • (1,0) dt  +  ∫0Δy F(a+Δx,b+t) • (0,1) dt
  		+ ∫0Δx F(a+Δx−t,b+Δy) • (−1,0) dt  +  ∫0Δy F(a,b+Δy−t) • (0,−1) dt
  	=	∫0Δx p(a+t,b) dt  +  ∫0Δy q(a+Δx,b+t) dt
  		−  ∫0Δx p(a+Δx−t,b+Δy) dt  −  ∫0Δy q(a,b+Δy−t) dt
  	=	∫0Δx p(a+t,b) dt  +  ∫0Δy q(a+Δx,b+t) dt
  		−  ∫0Δx p(a+t,b+Δy) dt  −  ∫0Δy q(a,b+t) dt

  In the last step, in the third integral we substitute t̃ = Δx − t,  dt̃ = −dt, so the limits of integration t = 0 to Δx become t̃ = Δx to 0; then we switch the limits of integration to re-acquire a minus sign, and finally we rename the variable t̃ to be t again. Similarly for the fourth integral.
  Now, for small h₁, h₂, we may approximate p(a+h₁,b+h₂)  ≈  p(a,b) + ^∂p⁄_∂x h₁ + ^∂p⁄_∂y h₂ , and similarly for q. Thus:

  ∮ F(c) • dc	≈	∫0Δx p(a,b) + ∂p⁄∂x t dt  +  ∫0Δy q(a,b) + ∂q⁄∂x Δx + ∂q⁄∂y t dt
  		−  ∫0Δx p(a,b) + ∂p⁄∂x t + ∂p⁄∂y Δy dt  −  ∫0Δy q(a,b) + ∂q⁄∂y t dt
  	=	− ∂p⁄∂y ΔxΔy  +  ∂q⁄∂x ΔxΔy,

  and dividing by the area ΔxΔy gives the desired formula ^∂q⁄_∂x − ^∂p⁄_∂y .
  This was not an exact computation, but the linear approximation is close enough to give the correct rate-of-circulation limit as Δx,Δy → 0. (To make this rigorous, we would need to analyze the error, which is quadratic in Δx and Δy, to make sure it cannot change the limit.)
• Moral of the story: A vector field F = (p,q) has a linear approximation given by the partial derivatives of its components:
  F(a+h₁, b+h₂)   ≈   F(a,b)  +  (^∂p⁄_∂x ,^∂q⁄_∂x) h₁  +  (^∂p⁄_∂y ,^∂q⁄_∂y) h₂ .
  However, this type of derivative is not very meaningful geometrically or physically.
  To get a more geometric derivative, we consider the circulation of F, and take the rate of circulation, curl F = ^∂q⁄_∂x − ^∂p⁄_∂y , a combination of two of the slope coefficients which indicates how strongly F circulates or rotates near a given point. (This is precisely the quantity which must vanish to make F conservative in a given region of the plane.)

• Double integral is volume below graph z = g(x,y), above planar domain D.
  For rectangle domain:
  D = [a,b]×[c,d] = {(x,y) with a ≤ x ≤ b , c ≤ y ≤ d},
  take sample points in x-interval and y-interval:
  a = x₀ < x₁ < ··· < x_n = b
  c = y₀ < y₁ < ··· < y_n = d.
  Sum up volumes of "french fries" above n×n tiny rectangles at (x_i,y_j):
  volume is: height g(x_i,y_j), times base area Δx_i Δy_j.
  ∬_D g(x,y) dx dy   =   lim_n→∞ ∑ _i,j=0 ⁿ⁻¹ g(x_i,y_j) Δx_i Δy_j .
  We can equally well write dy dx instead of dx dy.
• Cavalieri Principle: To compute volume, integrate areas of all slices perpendicular to an axis.
  Fubini Theorem: To compute double integral ∬_D g(x,y) dx dy,
  • Evaluate area of each slice perpendiuclar to y-axis: the inside integral G(y) = ∫_a^b g(x,y) dx with respect to x-variable, considering y as constant.
  • Integrate ∫_c^d G(y) dy along y-axis.
  • Equivalently, can switch to ∬_D g(x,y) dy dx, doing inside y-integral first.

  ∬_D g(x,y) dx dy  =  ∫_c^d (∫_a^b g(x,y) dx) dy  =  ∫_a^b (∫_c^d g(x,y) dy) dx.

• Example: Rectangular region D = [0,1]×[2,3] with 0 ≤ x ≤ 1 and 2 ≤ y ≤ 3, function g(x,y) = xy², we have:
  ∬_D xy² dx dy  =  ∫³₂ (∫¹₀ xy² dx) dy
   =  ∫³₂ [¹⁄₂x²y²]_x=0 ^x=1   dy  =  ∫³₂ ¹⁄₂y² dy  =  ¹⁹⁄₆ .
  ∬_D xy² dy dx  =  ∫¹₀ (∫³₂ xy² dy) dx
   =  ∫¹₀ [¹⁄₃xy³]_y=2 ^y=3   dy  =  ∫¹₀ ¹⁄₃(27x − 8x) dx  =  ¹⁹⁄₆ .

• Definition of ∬_D g(x,y) dx dy for general domain D:
  In Riemann sum, add only the sample points (x_i,y_j) ∈ D
• Simple domains: generalize (x,y) ∈ D = [a,b] × [c,d]
  x-simple D = {(x,y)  |  a ≤ x ≤ b , c(x) ≤ y ≤ d(x)}
  y-simple D = {(x,y)  |  c ≤ y ≤ d , a(y) ≤ x ≤ b(y)}
• Example: D triangle bounded by y = x , y = 1−x , y = 0; integrate g(x,y) = x²+y².
  Above x ∈ [0,1], y-floor is y = 0, but y-ceiling is different on left vs right.
  Split into x-simple regions D₁ = {(x,y)  |  0 ≤ x ≤ ½ , 0 ≤ y ≤ x } and D₂ = {(x,y)  |  ½ ≤ x ≤ 1 , 0 ≤ y ≤ 1−x }:
  ∬_D x²+y² dy dx   =   ∫ _x=0 ^x=1/2 ∫ _y=0 ^y=x  x²+y² dy dx  +  ∫ _x=1/2 ^x=1 ∫ _y=0 ^y=1−x  x²+y² dy dx
    =   ∫ _x=0 ^x=1/2 x³ + ⅓x³ dx  +  ∫ _x=1/2 ^x=1 x²(1−x) + ⅓(1−x)³ dx   =   ¹⁄₄₈ + ¹⁄₁₆   =   ¹⁄₁₂ .
  The other way is easier. To the right of y ∈ [0,½], x-floor is x = y and x-ceiling is x = 1−y,
  D = { (x,y)  |  0 ≤ y ≤ ½ , y ≤ x ≤ 1−y }:
  ∬_D x²+y² dx dy   =   ∫ _y=0 ^y=1/2 ∫ _x=y ^x=1−y  x²+y² dx dy
    =   ∫ _y=0 ^y=1/2 ⅓(1−y)³ + (1−y)y² − ⅓y³ − y³ dy   =   ¹⁄₁₂ .

• Example: D bounded between y = 2x² and y = x²−2x+3.
  Find points of intersection to determine a ≤ x ≤ b,  x²−2x+3 ≤ y ≤ 2x²
• Technique: Reversing the order of integration
  ∫ _x=a ^x=b ∫ _y=c(x) ^y=d(x) g(x,y) dy dx  ⇒  sketch x-simple D
   ⇒  convert to y-simple D: y-shadow c ≤ y ≤ d
        for given y, find x-floor and ceiling a(y) ≤ x ≤ b(y)
   ⇒  ∫ _y=c ^y=d ∫ _x=a(y) ^x=b(y) g(x,y) dx dy

Reading: [MT] Ch 5.1, 5.2, 5.3. Double integrals.
HW:

1. A bunch of double integrals, done both ways (dx dy and dy dx):
   [MT] Ch 5.1 p 269, Ex #1,2,3,4. Ch 5.2 p 282, Ex #1.
2. Sketch the solid region whose volume is computed by the given integral.
   1. ∬_R x+y² dx dy for R = [0,1]×[−1,1]
   2. ∬_R 1+xy dx dy for R = [−1,1]×[−1,1]

1. See [MT] answers, p 517, or evaluate (indefinite or definite) double integrals using Wolfram. Note that doing an integral as dx dy or as dy dx should give the same answer.

2a. The solid is under the graph z = x+y² and above the rectangle (x,y) ∈ [0,1]×[−1,1]. To visualize the graph, consider the "backbone" slice above the x-axis (y=0), which is just the line z = x; and the "ribs" slices perpendicular to the x-axis, above x = c, which are parabolas z = c+y² rising from the backbone above the x-axis. See the graph in Wolfram.

2b. The graph is a quadrant of a saddle surface, with upward parabola above y = x, downward parabola above y = −x, and two flat lines above the x and y axes.

• Circulation around a region splits as the region splits
  For D = D₁ ∪ D₂ with counterclockwise boundaries c, c₁, c₂, we have:
  
  ∮ F(c)•dc   =   ∮ F(c₁)•dc₁ + ∮ F(c₂)•dc₂
  
• Splitting D into a grid of n small boxes D  =  D₁ ∪ ··· ∪ D_n , each box D_i with corner point (x_i, y_i) and size ΔxΔy, has approximate circulation:
  ∮ F(c_i)•dc_i  ≈  curl F(x_i, y_i) ΔxΔy.
  Thus, the line integral around D splits as a sum over the n small boxes:
  ∮ F(c_i)•dc_i  =  ∑ⁿ_i=1 ∮ F(c_i)•dc_i  ≈  ∑ⁿ_i=1 curl F(x_i, y_i) ΔxΔy  ≈  ∫∫_D curl F(x,y) dx dy.

• The last term is a double integral ∫∫ g(x,y) dx dy of the function g(x,y) = curl F(x,y) over a region D, and the approximations become equalities as n → ∞. That is,
  ∫∫ g(x,y) dx dy   =   lim_n→∞ ∑ⁿ_i=1 g(x_i, y_i) ΔxΔy,
  where (x_i, y_i) are corner points of small boxes D_i of size ΔxΔy which roughly fill D.
• We have discovered another result analogous to the Fundamental Theorem of Calculus.
  Curl Theorem: (Fundamental Theorem for Curl) The total circulation of a vector field around a region is equal to the integral of the rate of circulation of the field within the region:
  ∮ F(c)•dc  =  ∫∫_D curl F(x,y) dx dy,
  where c is the counter-clockwise boundary curve of the region D.

HW 14: Continue WHW 5 below.

Example of the Curl Theorem. For the problems below, consider the vector field:
F(x,y)  =  (x+y)(−1,1)  =  (−x−y, x+y).
Let D be the region between the parabola curves y = x² and y = 2 − x², with counterclockwise boundary curve c.
1. Sketch the field F and the region D. Estimate the circulation ∮ F(c)•dc as positive, zero, or negative, and the same for curl F and its integral ∬_D curl F(x,y) dy dx.
2. Write the region in the form D = {(x,y) with a ≤ x ≤ b, c(x) ≤ y ≤ d(x)}. Use this to compute the double integral ∬_D curl F(x,y) dy dx in exact terms.
3. Parametrize the curve c in two pieces c₁(t), c₂(t). Use this to compute the circulation loop integral ∮ F(c)•dc. Hint: Both pieces must go counter-clockwise around D. Take x = t or x = 1−t, each for t ∈ [−1,1].
4. Compare the answers to #2 & #3. Why should they be equal? Do they agree with your estimates in #1?

1. F is zero on y = −x and flows along the lines y = −x + k, with increasing flow above and below y = −x. The region D looks like an eye, with left and right corners (−1,1), (1,1), top (0,2), bottom (0,0). The field flows mostly counter-clockwise around D, in roughly the same direction as the counter-clockwise boundary curve, meaning positive circulation. The field clearly spins counter-clockwise near every point, meaning positive curl with positive double integral. The Curl Theorem says these answers must be the same: the total circulation around D is equal to the integral of the rate of circulation inside D.

More quantitatively, the field vectors around the boundary have length about 1 or 2, somewhat oblique from the tangent direction, and the curve has length about 6 (like a unit circle), so the line integral might be about 5? Also, the field vectors grow at a moderate linear rate, so the curl is not large, and the region has area about 3 (like a unit circle), so we would expect a double integral around 5? These are not reliable estimates, but they give an idea of the magnitude and sign of the answers (which in fact must be exactly equal).

2. D = {(x,y) with −1 ≤ x ≤ 1 and x² ≤ y ≤ 2−x²}. Compute the double integral ∬_D curl F(x,y) dy dx as in the previous HW. Note: This integral would be harder to do in the opposite order ∬_D curl F(x,y) dx dy because D would have to be split into two pieces with different formulas. Answer: ¹⁶⁄₃ .

3. The bottom curve is c₁(t) = (t, t²) for −1 ≤ t ≤ 1, moving left-to-right under D (counter-clockwise). The top curve is c₂(t) = (−t, 2−t²) for −1 ≤ t ≤ 1, moving right-to-left from c₂(−1) = (1,1) to c₂(1) = (−1,1), so as to go counter-clockwise above D. Then ∮_c F = ∫_c1 F + ∫_c2 F. Answer: ¹⁶⁄₃ .

4. The Curl Theorem (Fund Thm for Curl) guarantees that the circulation of F around c (the boundary of D) is equal to the double integral of the rate of circulation inside D, meaning #2 and #3 must give the same answer. This should be consistent with the estimates from #1.

• Flux integral. The flux of a vector field F across a curve c is the net flow of F from the left to the right side of c, facing forward along the curve.
• By this we mean that c(t) = (x(t), y(t)) has velocity vector c'(t) = (x'(t), y'(t)), whose counterclockwise orthogonal c'(t)^⊥ = (−y'(t), x'(t)) points to the left side of c; and the right-side orthogonal n(t) = −c'(t)^⊥ = (y'(t), −x'(t)) is called the normal vector.
• To see how strongly the field F is flowing across the curve c at a given point, we take the dot product of F(c(t)) • n(t).
  Summing this over sample points c₀, . . . ,c_n on the curve c, and Δc_i = c_i+1−c_i ≈ c'(t_i) Δt_i, and taking Δt_i → 0, we obtain a new kind of line integral:
  Flux of F across c   =   ∫ F(c) • dn   =   ∫ F(c) • (−dc^⊥)   =   ∫¹₀ F(c(t)) • (−c'(t)^⊥) dt.

  We compute this in coordinates, with c(t) = (x(t),y(t)) for t ∈ [0,1]:

  ∫ F(c) • dn  =  ∫¹_t=0 F(x(t),y(t)) • (y'(t),−x'(t)) dt.

• The flux integral over a counterclockise loop c is denoted ∮ F(c) • dn, and computes the net outflow from the region enclosed by c.
• Divergence: For a vector field F on the xy-plane, we define its divergence at (x,y) as the rate of outflow of F across a small loop c near (x,y), relative to the area enclosed by c:
  div F(a,b)   =   lim_c→(x,y) ( ∮ F(c) • dn ) / area(c) .
  For example, we may take c(t) to be the curve which traces counterclockwise around a rectangle from (x,y) to (x+Δx, y) to (x+Δx, y+Δy) to (x, y+Δy) and back to (x,y), and dn(t) = −dc(t)^⊥ is the outward normal vector of the rectangle.
  The above limit can be computed to give a formula in terms of the partial derivatives of the components:
  F(x,y) = (p(x,y), q(x,y))
  
  div F(x,y)   =   ^∂p⁄_∂x + ^∂q⁄_∂y .
  If div F(x,y) = 0 everywhere, we say F is incompressible, modeling velocity of a fluid in which there is no net outflow from any region.
• Example: The field F(x,y) = (¹⁄_(1+x²), 0) could represent the velocity of the Red Cedar River flowing over the dam (the y-axis): the river gets faster where it is shallower.
  
  The unit box in red has larger inflow on the left than outflow on the right, with no outflow on top or bottom, so the flux integral ∮ F(c)•dn must be negative. We compute by adding up integrals for the four linear segments of c. The top and bottom normals dn are perpendicular to the field, so they give zero. The left and right contributions are:
  ∮ F(c) • dn   =   ∫¹_t=0 F(0,1−t)•(−1,0) dt + ∫¹_t=0 F(1,t)•(1,0) dt
    =   ∫¹_t=0 (1,0)•(−1,0) dt + ∫¹_t=0 (½,0)•(1,0) dt   =   −½.
  Similarly, a small box placed at (x,y) = (1,0) would have a stronger inflow across its left boundary and a weaker outflow across its right boundary, giving net inflow and negative divergence. We compute:
  div F(x,y) = ^∂⁄_∂x(¹⁄_(1+x²)) + ^∂⁄_∂y(0)  =  ^−2x⁄_(1+x²)² ,
  and we verify that div F(1, 0) = −½.

Reading: [MT] 4.4 p. 245−248; Ex 5, 7−10, p. 258.
HW:

1. Picturing flux. Let c be the counter-clockwise ellipse defined by 4x² + y² = 1. For each of the following vector fields F:
   • Sketch F, and visually estimate the flux ∮ F(c)•dn.
   • Compute the line integral exactly using coordinates, and compare to your estimate.
   The fields are:
   1. F(x,y) = (0, x)
   2. F(x,y) = (x, 0)
   3. The radial field F(x,y) = (x,y).
   4. A rotational field F(x,y) = g(r) (−y, x), where g(r) is any scalar function of the radius r = √(x² + y²). For example g(r) = r⁻², giving F(x,y)  =  (^−y⁄_(x²+y²) , ^x⁄_(x²+y²)).
2. Picturing divergence. For the same vector fields above:
   • From the sketch of F, visually estimate div F, the rate of outflow at each point (x,y).
   • For F(x,y) = (p(x,y), q(x,y)), compute div F = ^∂p⁄_∂x + ^∂q⁄_∂y, and compare to your estimate.

The ellipse curve is a squeezed circle: c(t) = (½cos(t), sin(t)), with tangent vector c'(t) = (−½sin(t), cos(t)) and normal vector n(t) = −c'(t)^⊥ = (cos(t), ½sin(t)).

1 & 2a. F(x,y) = (0,x) The flux across each point on the top half of the ellipse is opposite the flux at the corresponding point on the bottom half. This cancellation means the total flux is zero. This is verified by computing the flux integral:

∫₀^2π F(½cos(t), sin(t)) • (cos(t), ½sin(t)) dt  =  0.

Also, any small box has zero flow across the vertical sides, and bottom inflow equal to top outflow, so div F(x,y) = 0 everywhere. Compute: div F = ^∂⁄_∂x(0) + ^∂⁄_∂y(x) = 0.

1 & 2b. F(x,y) = (x,0)

The flux is outward (positive) at each point of the ellipse, so the flux integral is positive. Any small box with x > 0 has zero flow across the horizontal sides, and left inflow less than right outflow, so div F(x,y) > 0, and similarly positive for x = 0 and x < 0. Compute: div F = ^∂⁄_∂x(x) + ^∂⁄_∂y(0) = 1.

1 & 2c. F(x,y) = (x,y)

The flux is outward (positive) at each point of the ellipse, so the flux integral is positive. Any small polar box has zero flow across the radial sides, and inner-arc inflow a lot less than outer-arc outflow, so div F(x,y) > 0 everywhere. Compute: div F = ^∂⁄_∂x(x) + ^∂⁄_∂y(y) = 2.

1 & 2d. F(x,y) = sin(x²+y²) (−y,x)

The flow at each point on the left of the ellipse is opposite to the flow at the corresponding point on the right, so the net outflow is zero. Any small polar box has zero flow across the circular sides, and the inflow across one radial edge is equal to the outflow across the other radial edge, so div F(x,y) = 0 everywhere. Compute: div F  =  ^∂⁄_∂x(g(x²+y²)(−y)) + ^∂⁄_∂y(g(x²+y²)x)
 =  g'(x²+y²)(2x)(−y) + g'(x²+y²)(2y)(−x)  =  0

• Divergence Theorem (Green's Theorem for Flux): For a vector field F, the double integral of the rate of outflow of F over a region D is equal to the total outflow of F across the boundary of D:
  ∬_D div F dx dy   =   ∮ F(c) • dn .
  Here the closed curve c(t) = (x(t),y(t)) must not cross itself, and it must go counterclockwise enclosing D, a region with no holes or points where div F is undefined. Also recall that dn = (y(t),−x'(t)) dt is the outward-pointing normal vector at each boundary point.
  The intuitive justification is that the flux line integral on the right side can be split over a grid of subregions at (x_i,y_j), whose flux integrals cancel each other along their common boundaries, except along the outside boundary of R. As the grid gets finer and finer, each small line integral near (x_i,y_j) is approximated by div F(x_i,y_j) Δx Δy, and their sum approaches the double integral on the left side.
• Example: Consider F(x,y) = (¹⁄_(1+x²), 0), over the triangle region D, defined y-simply by 0 ≤ y ≤ 1 , 0 ≤ x ≤ 1−y. The double integral of the rate of outflow (the divergence of F) over the region is:
  ∬_D div F dx dy   =   ∫₀¹ ∫₀ ^{1−y −2x}⁄_(1+x²)² dx dy

    =   ∫¹₀ [¹⁄_1+x² |^1−y _x=0 dy   =   ∫¹₀ ¹⁄_1+(1−y)² − 1 dy   =   ^π⁄₄ − 1.

  The boundary of R consists of the line segments c₁(t) = (t,0), c₂(t) = (1−t,t), c₃(t) = (0,1−t), each for 0 ≤ t ≤ 1; and the total flux is the line integral:

  ∮ F(c) • dn   =   ∮ F(c₁) • dn₁ + ∮ F(c₂) • dn₂ + ∮ F(c₃) • dn₃
    =   0 + ^π⁄₄ − 1 .

  In detail, one of these is:

  ∮ F(c₂) • dn₂   =   ∫¹₀ F(1−t, t) • (−(1−t, t)')^⊥ dt
    =   ∫¹₀ (¹⁄_(1+t²), 0) • (−(−1,1)^⊥) dt   =   ∫¹₀ (¹⁄_(1+t²), 0) • (1,1) dt   =   ^π⁄₄ .
  Notice that n = (1,1) is the outward normal from the upper edge of the triangle.

Reading: [MT] 4.4 p. 245−248; Ex 5, 7−10, p. 258.
HW16 :

1. Consider the radial vector field with length inversely proportional to the distance from the center: F(x,y)  =  ¹⁄_(x²+y²) (x,y)  =  (^x⁄_(x²+y²) , ^y⁄_(x²+y²)).
   1. Verify that the length |F(x,y)| = 1/r. Sketch the vector field F, noting that it is undefined at the origin, and guess div F.
   2. Show that div F = 0 everywhere except at the origin, meaning F is incompressible. This means the divergence is all "concentrated in the center".
   3. Compute the flux integral (total outflow) ∮_c F • dn across the boundary curve c of the polar-box region D defined by 1 ≤ r ≤ 2, 0 ≤ θ ≤ ^π⁄₂. Verify that it is indeed zero: by the Divergence Theorem, the total flux out of D must equal the integral of the rate of flux within D:  ∬_D div F dx dy. But div F = 0.

1a. F(x,y)  =  (^x⁄_(x²+y²) , ^y⁄_(x²+y²)). A rough sketch, with the large vectors near the origin not show to scale:

For any small radial box, the radial edges have zero flow; and the inner circular edge has a shorter arc with longer inflowing F vectors, while the outer circular edge as a longer arc with shorter outflowing F vectors. Thus we may guess div F ≈ 0 everywhere (except at the origin, where there is a positive outflow from zero area, an infinite rate of outflow).

1b. Compute for radial field F(x,y) = g(x²+y²) (x,y) = ¹⁄_(x²+y²) (x,y):

div F  =  ^∂⁄_∂x(g(x²+y²)x) + ^∂⁄_∂y(g(x²+y²)y)
 =  g'(x²+y²)(2x²) + g(x²+y²) + g'(x²+y²)(2y²) + g(x²+y²)
 =  ^−2(x2+y2)⁄_(x²+y²)² + ^2(x2+y2)⁄_(x²+y²)²  =  0.

1c. The radial box consists of curves: c₁(t) = (t+1, 0)  ,  c₂(t) = (2 cos(^π⁄₂ t), 2 sin(^π⁄₂ t)),
c₃(t) = (0, 2−t)  ,  c₄(t) = (cos(^π⁄₂(1−t)), sin(^π⁄₂(1−t))), where all parameters run over 0 ≤ t ≤ 1. These have tangent vectors: c'₁(t) = (1,0)  ,  c'₂(t) = (−π sin(^π⁄₂ t), π cos(^π⁄₂ t)),
c'₃(t) = (0,−1)  ,  c'₄(t) = (^π⁄₂ sin(^π⁄₄(1−t)) , ^−π⁄₂ cos(^π⁄₄(1−t))). Since n(t) = −c'(t)^⊥, where −(x,y)^⊥ = (y,−x), we have: n₁(t) = (0,−1)  ,  n₂(t) = (π cos(^π⁄₂ t), π sin(^π⁄₂ t)),
n₃(t) = (−1,0)  ,  n₄(t) = (^−π⁄₂ cos(^π⁄₂(1−t)) , ^−π⁄₂ sin(^π⁄₂(1−t))). The flux integral is a sum of four line-integrals over c₁, c₂, c₃, c₄, for example: ∮ _c2 F(c₂(t)) • dn₂  =  ∫₀¹ F(2 cos(^π⁄₂ t), 2 sin(^π⁄₂ t)) • (π cos(^π⁄₂ t), π sin(^π⁄₂ t)) dt
 =  ∫₀¹ (¹⁄₂ cos(^π⁄₂ t), ¹⁄₂ sin(^π⁄₂ t)) • (π cos(^π⁄₂ t), π sin(^π⁄₂ t)) dt  =  ^π⁄₂ . Note this is positive because there is a net outflow across c₂: that is, F(c₂(t)) flows from left to right, facing along c'₂(t). Similarly, the integral for c₄ is ^−π⁄₂ , and the other two integrals are both zero, since F is orthogonal to the normal.

• Re-parametrization. We usually deal with a curve c by parametrizing it as c(t) = (x(t), y(t)), where the controlling variable t is called a parameter. In fact, any curve can be parametrized in many ways, traversing the same points at different rates.
• Example: Let c be the top half of a unit circle, oriented from right to left. We usually parametrize this by angle:
  c₁(t) = (cos(t), sin(t))   for   t ∈ [0,π].
  But we can traverse the same curve starting at (1,0), moving left by s to (x,0) = (1−s, 0), then going up by y = √(1−x²):
  c₂(s) = (1−s, √(1−(1−s)²)   for   s ∈ [0,2].
  Since these traverse the same points, every c₂(s) is equal to c₁(t) for some t depending on s, i.e. t = g(s). In fact, determining the radial angle t of the point c₂(s) gives:
  t = arccos(1−s) = g(s),     so     c₂(s) = c₁(g(s)).
  There is such a change-of-variable or substitution function g for any two parametrizations of the same curve.
• Chain Rule: For two parametrizations c₁(t) = c₂(s) with t = g(x), the velocity vectors are related by:
  c'₂(s) = c'₁(g(s)) g'(s).
  Here g'(s) is a scalar function multiplying the vector c'₁(g(s)).
  This can be proved from the single-variable Chain Rule by computing the two sides in coordinates.
• Independence of Parametrization: For any vector field F (not necessarily conservative) and a curve c with specified direction, the path integral ∫ F(c) • dc gives the same value for any parametrization of c.
  Proof: The single-variable substitution t = g(s), dt = g'(s) ds gives:
  ∫ F(c₁) • dc   =   ∫ F(c₁(t)) • c'₁(t) dt   =   ∫ F(c₁(g(s))) • c'₁(g(s)) g'(s) ds
  
    =   ∫ F(c₂(s)) • c'₂(s) ds   =   ∫ F(c₂) • dc₂ .

• Arclength of c(t) = (x(t),y(t)) for a ≤ t ≤ b:
  L  =  lim_n→∞ ∑ _i=0 ⁿ⁻¹  |Δc_i|  =  lim_n→∞ ∑ _i=0 ⁿ⁻¹  |c'(t_i)| Δt
  
   =  ∫_a^b |c'(t)| dt  =  ∫_a^b √(x'(t)²+y'(t)²) dt .
  Length is the integral of speed (magnitude of velocity).

HW 16½

1. Prove that the above arclength integral is the same for any two parametrizations c₂(s) = c₁(t) for t = g(s).
2. Give an integral formula for the length of the graph curve y = f(x), parametrized as c(t) = (t, f(t)).
3. Give an integral formula for the length of the polar graph curve r = f(θ), parametrized as c(t) = f(t) (cos(t), sin(t)).
4. Compute the length of the unit semi-circle using the parametrizations (cos(t), sin(t)) for t ∈ [0,π] and (s, √(1−s²)) for s ∈ [−1,1]. (Direction matters for path integrals, but not for arclength.) You must get the same answer π both times, but one computation is much easier.
   Hint: Make a change of variable to transform the hard one into the easy one.
5. Arclength.
   1. Find the arclength of the graph y = x² for 0 ≤ x ≤ 1. Hint: Recall the integral formula ∫√(a²+x²) dx = ^x⁄₂√(a²+x²) + ^a2⁄₂ log(x+√(a²+x²)) + C.
   2. Recall the cycloid curve c(t) = (t−sin(t), 1-cos(t)), the path of a point on the rim of a rolling wheel. Find the length of one arch of the curve, from t = 0 to t = 2π. Hint: Use a trig half-angle formula to evaluate the integral.
   3. Find the arclength of a snail-shell spiral r = e^θ.

• Function G : R² → R², G(u,v) = (g₂(u,v), g₂(u,v)) = (x(u,v), y(u,v))
  Transformation moves and stretches plane: picture u,v grid lines in x,y plane
• Linear G(u,v) = Shear(u,v) = (u+v, v), squashed grid
• Translation G(u,v) = Tran(u,v) = (u+½, v−⅓), shifted grid
• Polar coord map P(u,v) = (u cos(v), u sin(v)) or P(r,θ) = (r cos(θ), r sin(θ))
  Curved grid of circles and radial lines. Restrict to domain (r,θ) ∈ [0,∞) × [0,2π)
• Parametrize a region: R = half-circle x² + y² ≤ 4, x ≥ 0
  P(r,θ) for parameter region (r,θ) ∈ R* = [0,2]×[−^π⁄₂,^π⁄₂]
• Ellipse R, (x−1)² + 4y² = 1, define elliptical coords
  G(u,v) = (1 + u cos(v), ½u sin(v)) for parameter region (u,v) ∈ R* = [0,1]×[0,2π]
• Two sets of contours u = f(x,y) = c and v = g(x,y) = c form a curvy grid in xy-plane
  Parameters u, v are two measurements on a point to determine its location on the curvy grid
  Inverse mapping G⁻¹(x,y) = (f(x,y), g(x,y)) = (u,v).
  Given curvy rectangle D with boundary contours f(x,y) = a₁, f(x,y) = a₂, g(x,y) = b₁, g(x,y) = b₂
  Take parameters u = f(x,y), v = g(x,y), invert to get x,y in terms of u,v
        G⁻¹(x,y) = (f(x,y), g(x,y)) = (u,v)   ⇒   G(u,v) = (x(u,v), y(u,v))
        D* = [a₁, a₂]×[b₁, b₂] with G(D*) = D
• Review definition of deriv: g(t) ≈ g(a) + g'(a)(t−a)  ⇔  g'(a) ≈ ^{(g(t)−g(a))}⁄_(t−a)
        g(a+h) ≈ g(a) + g'(a)h  ⇔  g'(a) ≈ ^{(g(a+h)−g(a))}⁄_h
• For g(v) = g(x,y), difference quotient makes no sense
  Lin approx g(a+h) ≈ g(a) + ∇g(a)•h = g(a) + Dg_a(h)
  Matrix of linear approx [Dg_a] = [∇g(a)] = [^∂g⁄_∂x(a)  ^∂g⁄_∂y(a)]
• Deriv of G(u,v) = (g₁(u,v), g₂(u,v)) = (x,y): curved grid looks linear close up
  Linear approx:   G(a+h)  ≈  G(a) + DG_a(h)
        (g₁(a+h), g₂(a+h))  ≈ (g₁(a), g₂(a)) + (∇g₁(a)•h, ∇g₂(a)•h)
  In matrix notation:

  g1(a+h) g2(a+h)

  ≈

  g1(a) + ∇g1(a)•h g2(a) + ∇g2(a)•h

  =

  g1(a) g2(a)

  +

  ∇g1(a)   ∇g2(a)

  •

  [h]

  .

  Jacobian matrix in terms of rows (gradients):

  [DGa]

  =

  ∇g1(a)   ∇g2(a)

  =

  ∂g1⁄∂u(a)   ∂g1⁄∂v(a)  ∂g2⁄∂u(a)   ∂g2⁄∂v(a)

  =

  ∂x⁄∂u(a,b)   ∂x⁄∂v(a,b)  ∂y⁄∂u(a,b)   ∂y⁄∂v(a,b)

  Jacobian matrix in terms of columns (grid curve tangent vectors):

  [DG_a]  =  [^∂G⁄_∂u(a)  ^∂G⁄_∂v(a)]

• Ex: Jacobian of polar coord mapping P(r,θ) = (r cos(θ), r sin(θ))

  [DP(r,θ)]

  =

  ∂⁄∂r(r cos(θ))   ∂⁄∂θ(r cos(θ))  ∂⁄∂r(r sin(θ))   ∂⁄∂θ(r sin(θ))

  =

  cos(θ)   −r sin(θ)  sin(θ)    r cos(θ)

  Parameter plane points (r,θ) near a =(3,^π⁄₄) can be written (r,θ) = a + h = (3+Δr, ^π⁄₄+Δθ), where h = (Δr, Δθ) is a small increment, and we approximate: P(r,θ)  ≈  P(3,^π⁄₄) + DP_(3,^π⁄4)(Δr, Δθ)

  	=	3 cos(π⁄4)    3 sin(π⁄4)	+	cos(π⁄4)   −3 sin(π⁄4)  sin(π⁄4)    3 cos(π⁄4)	•	Δr Δθ

  Lin approx of polar grid near (r_o,θ_o):   DP_(ro,θo) = Rot_θo∘ Dil_(1,ro)

HW 17:

1. Parametrize a triangle
   1. Consider a (u,v)-grid in the (x,y)-plane R² whose u-lines (red) are parallel to the vector (2,1), and whose v-lines (blue) are parallel to (1,3).
      
      Find a (linear) mapping L(u,v) = (x(u,v), y(u,v)) which takes the standard grid in the (u,v)-plane to the above grid in the (x,y)-plane.
   2. Consider the triangular region R (light blue) with vertices (0,0), (2,1), (1,3). Find a simple parameter region R* in the (u,v) plane which is taken to R by the mapping L. This gives a parametrization of R. (Here R* should be a simple triangle easy to describe in terms of inequalities such as u ≥ 0 and v ≤ ℓ(u) for some function ℓ.)
   3. Find the inverse linear mapping L⁻¹(x,y) = (u(x,y), v(x,y)). Hint: Solve for u,v in terms of x,y in the system of equations x = x(u,v), y = y(u,v). Equivalently, find the inverse matrix [L]⁻¹.
   4. Now let R be any triangular region in the (x,y) plane with vertices a, b, c. Generalize the above exercise to define an affine linear function A(u,v) and a parameter region R* which parametrize R; recall that affine means constant + linear. (It may be convenient to define A(u,v) in terms of the vectors b−a, c−a, rather than in coordinates.)
2. Parametrizing rotated regions
   1. Parametrize the region R bounded by the ellipse ^x2⁄₄ + ^y2⁄₉ = 1. That is, find a function F(u,v) which takes some simple (u,v)-region R* to R. Hint: The extreme points of the ellipse are (±2,0) and (0,±3). Stretch the x and y coordinates of the usual polar parametrization of the unit disk to get the correct height and width.
   2. Now parametrize the same ellipse, but rotated 45° and with its center shifted to (1,2). Also find the inverse mapping. Hint: Take the the previous parametrization mapping, then compose it with rotation and translation mappings.
3. With colored pens, plot straight grid lines in the (r,θ) plane corresponding to r = 0, 1, 2, 3, and θ = 0, ^π⁄₄, ^π⁄₂, ^3π⁄₄, . . . , ^7π⁄₄, and the corresponding grid curves in the (x,y) plane, under the polar mapping
   P(r,θ)  =  (x(r,θ), y(r,θ))  =  (r cos(θ), r sin(θ)).
   Each grid curve is a parametrized curve in the (x,y)-plane, whose parameter is either r or θ. For example, for r = 2, the (r,θ)-gridline (2,θ) is taken to the curve c(θ) = P(2,θ) = (2 cos(θ), 2 sin(θ)).
4. Compute the derivative of P at the point (r,θ) = (2,^π⁄₄), the Jacobian matrix:

   [DP(2,π⁄4)]

   =

   ∂x⁄∂r(2,π⁄4)   ∂x⁄∂θ(2,π⁄4)  ∂y⁄∂r(2,π⁄4)   ∂y⁄∂θ(2,π⁄4)

   Note that once we substitute (r,θ) = (2,^π⁄₄), this becomes a matrix of explicit constants, with no variables.
5. Consider the affine approximation to P(r,θ) near (r,θ) = (2,^π⁄₄):
   P(2+Δr, ^π⁄₄+Δθ)  ≈  P(2,^π⁄₄) + DP_(2,^π⁄4)(Δr, Δθ),
   or equivalently:
   P(r,θ)  ≈  P(2,^π⁄₄) + DP_(2,^π⁄4)(r−2, θ−^π⁄₄).
   Here DP_(2,^π⁄4) is the linear mapping defined by the Jacobian matrix, and Δr, Δθ are small increments of r, θ.
   Problem: Take the colored (r,θ) grid lines above, and map them to grid lines in the (x,y)-plane under the affine linear approximation. (Recall that a parametrized line c(t) = a + tv starts at a and moves in direction v.) Note how these lines approximate the polar grid curves near (r,θ) = (2,^π⁄₄).
6. Inversion mapping
   1. Let Inv : R² → R² be the mapping which takes a non-zero vector (u,v) to a vector (x,y) with the same direction and the reciprocal radius: that is, Inv turns the unit circle "inside out" so that its interior is stretched to cover the whole plane outside the circle, and region outside the circle is squashed inside. Find a vector formula for Inv(v), as well as a coordinate formula Inv(u,v) = (x,y).
   2. Sketch the Inv(u,v)-grid produced by the above mapping in the (x,y)-plane. Hint: The integer grid lines in the (u,v) plane are all taken to small circles inside the unit circle. Since each (u,v)-grid line goes to infinity in the (u,v)-plane, its image goes to the origin in the (x,y)-plane. These are called Apollonian circles or a parabolic pencil of circles.
   3. Find the inverse mapping Inv⁻¹(x,y). Hint: Duh!
   4. Extra Credit: Inverting familiar curves in the (u,v) plane gives interesting new curves in the (x,y) plane. The inversion of the parabola v = u² − 1 is called the cardioid: you can parametrize it as c(u) = (x(u), y(u)) = Inv(u, u²−1). Similarly, the inversion of the hyperbola u² − v² = 1 is called the lemniscate of Bernoulli. Explore these curves, and say as much as you can about them, showing how the inversion definition is equivalent to the usual ones you can find online. For example, find (x,y) equations defining these curves, or polar coordinate equations r = f(θ). Does the inversion mapping give insight on their properties?

1a. We just need the (u,v) coordinate axis vectors to map as L(i) = (2,1) and L(j) = (1,3), and the rest of the grid follows. Thus L(u,v) = (2,1)u + (1,3)v = (2u+v, u+3v). Note that the matrix of L is [

2	1
1	3

].

1b. The parameter region is the standard triangle with vertices (u,v) = (0,0), (1,0), (0,1). That is, R* = {(u,v) with 0 ≤ u and 0 ≤ v ≤ 1−u} = {(u,v) with u ≥ 0, v ≥ 0, u+v ≤ 1}.

1c.   [ L ]⁻¹  =  ¹⁄₅ [

3	−1
−1	2

], so:

L⁻¹   =   (³⁄₅ x − ¹⁄₅ y , −¹⁄₅ x + ²⁄₅ y).

1d. A(u,v) = a + (b−a)u + (c−a)v will take the standard (u,v) grid to the grid generated by the edge-vectors b−a and c−a, with the origin taken to a. Use the previous parameter region R* = {(u,v) with u ≥ 0, v ≥ 0, u+v ≤ 1}.

2a. F(r,θ) = (¹⁄₂ r cos(θ), ¹⁄₃ r sin(θ)) over the region (r,θ) ∈ R* = [0,1]×[0,2π].

2b. The rotation map Rot_^π⁄4 is defined by a rotation matrix which works out to Rot(x,y) = ¹⁄_√2(x−y, x+y). Applying this to the output of the previous F(r,θ), then translating by (1,2), gives:

G(r,θ)  =  ( 1 + ^r⁄_2√2 cos(θ) − ^r⁄_3√2 sin(θ) , 2 + ^r⁄_2√2 cos(θ) + ^r⁄_3√2 sin(θ) ).

Here we should think of r as measuring an "elliptically scaled" radius measured from the center of the tilted, translated ellipse; and θ as measuring a squashed "elliptical angle".

3. How P maps the (r,θ)-gridlines to the (x,y)-plane:

P →

For example, the (r,θ)-gridline (2,θ) maps to the (x,y)-plane as the grid curve: c(θ) = P(2,θ) = (2 cos(θ), 2 sin(θ)), a circle of radius 2.

4. Specializing the Jacobian matrix in the above Notes to the point (r,θ) = (2,^π⁄₄) gives:

[DP(2,π⁄4)]

=

cos(π⁄4)   −2 sin(π⁄4)  sin(π⁄4)    2 cos(π⁄4)

=

√2⁄2   −√2 √2⁄2     √2

5. The affine approximation function to P near the base point (r,θ) = (2,^π⁄₄) is:

A(r, θ)   =   P(2,^π⁄₄) + DP_(2,^π⁄4)(r-2, θ−^π⁄₄)   =   (√2, √2) + (^√2⁄₂(r-2) − √2(θ−^π⁄₄) , ^√2⁄₂(r-2) + √2(θ−^π⁄₄)).

The corresponding gridlines are c(θ) = A(2,θ), A(3,θ), and parallel lines; and c(r) = A(r,^π⁄₄), A(r,^π⁄₈), and parallel lines:

The faint dotted lines are the polar grid curves. These two grids do indeed approach each other very close to the base point, on a scale where the grid curves look straight.

6a. The vector parallel to v = (u,v), having length ¹⁄_|v|, is F(v) = ^v⁄_|v|², which means Inv(u,v) = ( ^u⁄_u²+v² , ^v⁄_u²+v² ).

6b. To see the transformation of the (u,v) grid under the mapping Inv(u,v), apply a parametric curve plotter successively to c(t) = Inv(1,t), Inv(2,t), Inv(3,t) gives red circles labeled u = 1,2,3 with diameter 1, ¹⁄₂, ¹⁄₃, all passing through the origin and with centers on the y-axis. Similarly for blue circles c(t) = Inv(t,1), Inv(t,2), Inv(t,3) centered on the x-axis.

Note that each red circle crosses each blue circle at right angles, like the standard grid in the (u,v)-plane. Thus, while Inv(u,v) turns straight edges to circles, it preserves all corner angles. This is called a conformal mapping.

6c. The mapping Inv : R²−{0} → R²−{0} is clearly its own inverse, since Inv(Inv(v)) = v. If we think of the first R² as the (u,v)-plane and the second as the (x,y)-plane, then we may write the inverse mapping as Inv(x,y) = ( ^x⁄_x²+y² , ^y⁄_x²+y² ).

• Complex field C = R[i]: closure, reciprocals
• Geometry of multiplication, homothety matrix
• Complex analytic function: Df_α = cpx mult  ⇔  Cauchy-Riemann eqns  ⇔  conformal mapping
• Conjugate-analytic vect flds F(x,y) = f(x−iy)  ⇒  curl, div = 0

• Area stretching factor for linear L(u,v) is determinant
  Defined by L(1,0) = u = (a,b) , L(0,1) = v = (c,d).
  Stretching factor is area of parallelogram D with edges u,v
  Area  =  base×height  =  |u| |v| sin θ(u,v)
         =  |u^⊥| |v| cos θ(u^⊥,v)  =  u^⊥•v  =  −bc+ad  =  ad−bc for u^⊥  =  (−b,a)

  =  det[L]  =  det(u,v)  =

  det

  [  a   c  b d ]

  =  ad − bc

  Positive area if v counterclockwise from u, negative area if clockwise
  Properties of det(u,v): bilinear, alternating det(u,v) = −det(v,u) ⇒ formula
• Substitution Rule for 1-variable integrals: x = g(u)
  Local stretching factor: dx = g'(u) du = ^dx⁄_du du
        ∫ f(x) dx = ∫ f(g(u)) g'(u) du
• Substitution Rule for G(u,v) = (x,y), bijective G : D* → D
  Assume G is orientation-preserving: det[DG] ≥ 0 for all (u,v)
  Local stretching factor is det of Jacobian matrix of derivatives
    ∬_D f(x,y) dx dy for  =  ∬_D* f(x(u,v),y(u,v)) det[DG_(u,v)] du dv.
  Leibnitz notation: det[DG_(u,v)]  =  ^∂(x,y)⁄_∂(u,v)
• Tradeoff in using substitution (x,y) = G(u,v):
        simplify region (x,y) ∈ D  ⇒  (u,v) ∈ D*
        but complicate integrand f(x,y)  ⇒  f(x(u,v),y(u,v)) |^∂(x,y)⁄_∂(u,v)|
• Ex: Polar P(r,θ) = (x,y) to parametrize roundish regions
  Jacobian determinant:

  det[DP(r,θ)]	=   [∂P⁄∂r | ∂P⁄∂θ]	=	det	∇x(r,θ) ∇y(r,θ)	=	det	cos(θ) −r sin(θ) sin(θ) r cos(θ)

   =  r cos²(θ) + r sin²(θ)  =  r
  Compute volume under surface of revolution z = r² = x² + y²
  for (x,y) ∈ D = disk of radius 1, parametrized by (r,θ) ∈ D* = [0,1]×[0,2π].
  ∬_D x²+y² dx dy  =  ∬_D* ( (r cos(θ))² + (r sin(θ))² ) |det[DP_(r,θ)]| dr dθ
   =  ∫^2π₀ ∫¹₀ r² r dr dθ  =  ∫^2π₀ ¹⁄₄ dθ  =  ¹⁄₄ 2π  =  ^π⁄₂ .
  Note: This is exactly half the volume of the enclosing cylinder under z = 1.

Reading: [MT] Ch 6.2, including odd exercises p 326 (answers p 521).
HW:

1. Use an appropriate parametrization to find formulas for the following quantities:
   1. The volume under an arbitrary surface of revolution z = f(r) = f(√(x²+y²)) and above a ring-shaped region with a ≤ r ≤ b. You probably saw the resulting formula in Calculus II: what was it called?
   2. The integral of an arbitrary f(x,y) over the elliptical region D defined by (x−1)² + ^y2⁄₄ ≤ 1
   3. The integral of f(x,y) over the triangle with vertices a = (1,1), b = (3,2), c = (2,4).
2. Gaussian integral   I = ∫ _−∞ ^∞ e^−x2 dx. The integrand e^−x2 is the or bell-curve or normal distribution which gives the relative probability of a random variable x with expectation value zero. (For example, x could be the height of a randomly chosen woman, minus the average height.)
   To make this a true probability distribution, we must scale it so that the total probability (the integral) is equal to one. That is, we must divide by the value of the integral I, but this is difficult to compute since e^−x2 has no elementary anti-derivative. We outline an amazing solution using double integrals and change of coordinates.
   1. Compute the double integral of e^−(x2+y2) over the disk x² + y² ≤ R², for a fixed radius R, and find the limit as R → ∞. Use the polar coordinate substitution: ∬_D f(x,y) dx dy  =  ∬_D* f(r cos θ, r sin θ) r dθ dr.
   2. Show that the integral of e^−(x2+y2) over the rectangle [−R,R]×[−R,R] is equal to ( ∫ _−R ^R e^−x2)² dx = I².
   3. Compare the integral in (a) with that in (b), letting R → ∞, and compute the original integral I.
3. Vector geometry of the determinant
   • Consider vectors u = (a,b), v = (c,d), forming an angle from u to v given by α with −π ≤ α ≤ π (positive meaning counterclockwise). Let β be the angle from v to u^⊥, the 90° counterclockwise rotation of u.
   • Let D be parallelogram with vertices 0, u, v, u+v.
   • Let area(u,v) be the signed area of D, counted positively if α ≥ 0, and negatively if α ≤ 0.
   Problem: Considering |u| as the base length of D, show that the height of D is |v| sin(α) = |v| cos(β), and hence area(u,v) = u^⊥•v = ad−bc.
4. The following properties of the area function are easy to see geometrically:
   • Bilinear:   area(au₁+bu₂, v)  =  a area(u₁,v) + b area(u₂,v)
     area(u, cv₁+dv₂)  =  c area(u,v₁) + d area(u,v₂)
   • Skew-symmetric:   area(u,v)  =  −area(v,u)
   • Unit square:   area(i,j) = 1, where i = (1,0) and j = (0,1)
   Problem: Assuming the above properties, show that area(u,v) = det(u,v) = ad−bc for u = (a,b), v = (c,d).
5. p. 348 #15: Use an appropriate change of variables to compute the integral ∬_B exp(^(y−x)⁄_(y+x)) dx dy over the triangle B with vertices (0,0), (1,0), (0,1).
   1. Consider the change of variables transformation (s,t) = T(x,y) = (y−x, y+x). Find its matrix [T], and verify that:

      [

      s t

      ]

      =  [T] •

      [

      x y

      ]

   2. Find the image region B* = T(B): this is a triangle with vertices T(0,0), T(1,0), T(0,1).
   3. Given (s,t) = (y−x, y+x), solve the two linear equations to get x,y in terms of s,t. Use this to write the inverse transformation U = T⁻¹ with (x,y) = U(s,t). Write the matrix [U] and verify that [U]•[T] = I, the identity matrix. Also B = U(B*).
   4. Write the Change of Variables formula for our integral, transforming it by U to (s,t) variables over B*.
   5. Evaluate the integral.
6. Routine practice from M&T.
   1. p. 304 #15(a,b). Do with rectangular coordinates, then again with polar.
   2. p. 305 #25. Do with rectangular coordinates, then again with coordinates (s,t) defined by (x,y) = s (1,0) + t (2,1).
   3. p. 305 #27, 37.

1. Use the polar parametrization P(r,θ), having Jacobian determinant det[DP] = r, and which maps the parameter region D* = [a,b]×[0,2π] to the ring-region D = {(x,y) | a² ≤ x²+y² ≤ b²}. The volume is:

Vol  =  ∬_D f(√(x²+y²)) dx dy  =  ∫_a^b ∫₀^2π f(r) r dθ dr  =  ∫_a^b 2πr f(r) dr.

This is the familiar "Shell Method Formula" for solids of revolution, usually justified by adding up thin cylindrical shells having perimeter 2πr, height f(r), and thickness dr.

1b. The region D is an ellipse with center (1,0), height 1, and width 2, so it is parametrized by a shift and scaling of polar coordinates: E(u,v) = (x,y) = (1 + u cos(v), 2u sin(v)), over the parameter region D* = [0,1]×[0,2π]. This has Jacobian determinant:

det[DE(u,v)]	=	det	∇x(u,v) ∇y(u,v)	=	det	cos(v) −u sin(v) 2sin(v) 2u cos(v)

 =  2u cos²(v) + 2u sin²(v)  =  2u

Thus we can compute the integral:

∬_D f(x,y) dx dy  =  ∫₀¹ ∫₀^2π f(1 + u cos(v), 2u sin(v)) 2u dv du.

1c. The given triangle D is half of a parallelogram with edge-vectors u = b−a = (2,1) and v = c−a = (1,3). We parametrize D by the affine linear mapping A(u,v) = a + L(u,v) with L(i) = u, L(j) = v, so that A(u,v) = (1+2u+v, 1+u+3v), over the triangular parameter region D* = {(u,v) | u,v ≥ 0, u+v ≤ 1}. The Jacobian determinant is:

det[DA(u,v)]  =  det[L]  =  det  2   1  1 3  =  (2)(3)−(1)(1)  =  5.

That is, the mapping A(u,v) expands all areas by a constant factor of 5. Thus we can compute the integral:

∬_D f(x,y) dx dy  =  ∫₀¹ ∫₀^1−v f(1+2u+v, 1+u+3v) 5 du dv.

2. Solution: See M&T p. 332.

3.We have the vectors u = (a,b), v = (c,d), the projection p of v on u, and the projection q of v on u^⊥.

By elementary trigonometry, the "height" vector has length: |q| = |v| sin(α) = |v| cos(β). Hence:

base×height = |u| |q| = |u^⊥| |v| cos(β) = u^⊥•v = (−b,a)•(c,d) = ad−bc.

4. Use the linearity properties to expand det(u,v) = det(a(1,0)+b(0,1), c(1,0)+d(0,1)), then use skew-symmetry and unit-square to evaluate the remaining determinants, getting ad(1) + ac(0) + bd(0) + bc(−1). Note that skew-symmetry implies det(u,u) = −det(u,u), so det(u,u) = 0.

5a.   [T]  =

−1  1    1 1

5b. B* is the triangle with vertices (s,t) = (0,0), (−1,1), (1,1), given by 0 ≤ t ≤ 1,  −t ≤ s ≤ t.

5c. (x,y) = (−½s + ½t , ½s + ½t).   [U]  =  [T]−1  =

−½  ½    ½ ½

This is indeed the inverse of [T], since:   [U] • [T]  =

−½  ½    ½ ½

•

−1  1    1 1

=

1   0  0 1

.

5d,e. We have U(B*) = B, one-to-one and onto. Since U is linear, it is its own derivative at every point: DU_(s,t) = U, and |^∂(x,y)⁄_∂(s,t)| = |det[U]| = ½ . (Note the det is actually −½, meaning the mapping U is orientation-reversing, so we have to take the absolute value.) Also y−x = s, y+x = t. Thus:

∬_B exp(^(y−x)⁄_(y+x)) dx dy  =  ∬_B* exp(^s⁄_t)  |^∂(x,y)⁄_∂(s,t)| ds dt  =  ¹⁄₂ ∫¹₀ ∫^t_−t e^s/t ds dt
 =  ¹⁄₂ ∫¹₀ ∫^t_−t e^s/t ds dt  =  ¹⁄₄ (e−e⁻¹).

6. See [MT] pp. 519−520

• Vector x = (x₁, . . . , x_n) ∈ Rⁿ
• Dot product u•v = (u₁, . . . ,u_n)•(v₁, . . . , v_n) = u₁v₁ + ··· + u_nv_n
  Geometrically: u•v = |u| |v| cos θ_uv
• Linear subspace V ⊂ Rⁿ or R³, k-dimensional with basis {v₁, . . . , v_k}
  Parametrize: V = { v = L(t₁, . . . , t_k)  |  t₁, . . . , t_k ∈ R },
        for linear function L : R^k → Rⁿ,   L(t₁, . . . , t_k) = t₁v₁ + ··· + t_kv_k
  Equations: V = { v ∈ Rⁿ  |  q₁•v = ··· = q_n−k•v = 0 }
        for orthogonal vectors q₁, . . . , q_n−k ⊥ V
• Affine subspace A = V + a ⊂ Rⁿ or R³ for constant vector a
  Parametrize by function P(t₁, . . . , t_k) = L(t₁, . . . , t_k) + a
  Equations: A = { v ∈ Rⁿ  |  q₁•(v−a) = ··· = q_n−k•(v−a) = 0 }
• Linear function ℓ: R³ → R
  ℓ(x,y,z)  =  ax+by+cz  =  (a,b,c)•(x,y,z),   ℓ(v)  =  q•v  for  q = (a,b,c)
  Picture ℓ(v) as height of v in direction q, scaled by |q|
  Height ℓ(v) = 0 defines plane orthogonal to q, through origin
  Other level surfaces ℓ(v) = c define affine planes orthogonal to q
  Actual height = (length of projection of v onto direction q) = ^q⁄_|q| • v
• Determinant det(u,v,w) = signed volume of parallelopiped with edges from the origin parallel to u,v,w.
  Sign is positive (+) if u,v,w form a right-handed coordinate system
        negative (−) if they form a left-handed one.
• Properties: det(u,v,w) = −det(v,u,w) = −det(u,w,v)
        det(su+s'u',v,w) = s det(u,v,w) + s' det(u',v,w), similarly for variables v, w
        det(i,j,k) = 1; also det(i,i,j) = 0, etc.
• Formula for determinant:

  det

  u1  u2 u3  v1  v2 v3  w1  w2 w3

  =

  u1v2w3 − u1v3w2 − u2v1w3 + u2v3w1 + u3v1w2 − u3v2w1

  =

  ∑σ sgn(σ) uσ(1)vσ(2)wσ(3)

  ,

  where σ runs over all 3! = 6 permutation functions σ : {1,2,3} → {1,2,3}.
  We can denote a permutation by listing σ(1)σ(2)σ(3). For example, σ(1) = 3, σ(2) = 2, σ(3) = 1 is denoted σ = 321
• The sign of a permutation sgn(σ) equals +1 if σ is obtained from the identity permutation ε(i) = i by switching an even number of pairs of outputs; and sgn(σ) equals −1 if it is obtained by switching an odd number of pairs of outputs.
  For example the permutation σ = 321 is obtained from ε = 123 by switching the single pair of outputs 1 and 3, so sgn(σ) = −1; or switch 12, then 23, then 12, again sgn(σ) = −1.
• Cross product on vectors in R³ only (not Rⁿ), giving vec × vec = vec
  Geometrically: u×v = q, the vector which produces the lin fun ℓ(w) = det(u,v,w) = q•w
        Given u, v, the cross product u×v is the unique vector such that for all w.:
  det(u,v,w)  =  (u×v) • w .

• Formula for u×v

  ℓ(x,y,z) = det

  u1  u2 u3  v1  v2 v3  x   y   z

  =

  u1v2z − u1v3y − u2v1z + u2v3x + u3v1y − u3v2x

  =   (u2v3− u3v2) x  +  (−u1v3+u3v1) y  +  (u1v2− u2v1) z

  =

  q • (x,y,z)

  q  =  u×v  =  (u₁,u₂,u₃) × (v₁,v₂,v₃)  =  (u₂v₃− u₃v₂ , −u₁v₃+u₃v₁ , u₁v₂− u₂v₁)

• Determinant cofactor expansion

  u × v

  =

  det

  u1  u2 u3  v1  v2 v3   i   j k

  =

  det

  u2  u3  v2  v3

  i

  −

  det

  u1  u3  v1  v3

  j

  +

  det

  u1  u2  v1  v2

  k

• Properties of u×v come from properties of det
        u×v orthogonal to both u,v:  (u×v)•u = (u×v)•v = 0
        u, v, u×v form a right-handed basis
        anti-commutative:   u×v = −v×u,   u×u = 0
        distributive:   (u+u') × v  =  u×v + u'×v
        linear:   (au)×v  =  u×(av)  =  a u×v
• Length |u×v| = area of parallelogram spanned by u,v in R³ = |u| |v| sin(θ_uv)
  Proof:   Geometrically, volume(u,v,w)  =  (base area of u,v parallelogram) (height of w perp to u & v).
  Algebraically, volume(u,v,w)  =  det(u,v,w)  =  (u×v)•w  =  |u×v|   ^u×v⁄_|u×v|•w
         =  |u×v|   (height of w along direction q). Cancelling (height of w), we get area(u,v) = |u×v|.

Reading: [MT] 1.1, 1.2. Cross Product: 1.3, including odd exercises p. 49 (answers p. 494).
HW: Extra Credit HW due Thu 10/31.

1. Consider the affine plane A ⊂ R³ containing the points a = (0,1,1), b = (1,1,2), c = (1,2,0). (As usual, when we define a point by a vector, we mean the endpoint of the vector starting at the origin.)
   1. Find two vectors parallel to the plane, for example from a to b, and from a to c.
   2. Write a parametric formula P(s,t) = (x(s,t), y(s,t), z(s,t)) which outputs the points of the plane as we move s,t ∈ R. Hint: Start at a and move along the two directions from part (a).
   3. Find an orthogonal vector q with q•(v−a) = 0 for v ∈ A. Write a defining linear equation satisfied by all v = (x,y,z) ∈ A, in the form ax + by + cz = d.
   4. Determine the distance from the point r = (3,4,5) to the plane. Hint: The distance is the difference in height with respect to the unit orthogonal vector ^q⁄_|q|.
2. Consider the affine line C ⊂ R³ through the points a = (0,1,1) and b = (1,1,2) from the previous problem.
   1. Find a parametric form c(t) = (x(t), y(t), z(t)) for C.
   2. Find two vectors q₁, q₂ which are orthogonal to the direction of C. Write two defining equations satisfied by all v = (x,y,z) ∈ C.
3. For each ordered triple of vectors u, v, w:
   • Sketch the vectors and the parallelopiped they span.
   • Determine whether they form a right-handed triple: that is can you point your right index finger along u, middle finger along v, thumb along w, so that the thumb is not bent behind or through the plane of the other two fingers.
   • Determine the signed volume of the parallelopiped by evaluating the determinant det(u, v, w). Verify that a right-handed triple has positive det, a left-handed one has negative det.
   The triples are:
   1. u = (1,1,0), v = (1,0,1), w = (1,1,2)
   2. u = (1,0,1), v = (1,1,0), w = (1,1,2)
   3. u = (1,1,0), v = (1,0,1), w = (2,1,1)
4. Again consider the affine plane A through the points a = (0,1,1), b = (1,1,2), c = (1,2,0).
   1. Re-compute the vector q orthogonal to A as the cross product: q  =  (b−a)×(c−a)
   2. For a general point w = (x,y,z) ∈ A, expand the vector equation q•(w−a) = 0, or equivalently q•w = q•a, to again get the explicit coordinate equation ax + by + cz = d defining A.
5. Again consider the vectors u = (1,1,0), v = (1,0,1), w = (1,1,2).
   1. Compute the cross product q = u×v.
   2. Compute the volume of the u, v, w parallelopiped by evaluating the triple product (u×v)•w = q•w = det(u,v,w).
6. The cross product u×v can only be defined for three dimensional vectors, but determinant exists in any dimension Rⁿ, and the formula
   det(u₁, . . . , u_n−1, x)  =  ℓ(x)  =  q • x   for all x ∈ Rⁿ.
   defines a unique vector q, which we may think of as a "product" of n−1 vectors: q = u₁×···×u_n−1 .
   1. For n = 2, this operation acts on a single vector u, giving u×. What familiar operation is this?
   2. Find a cofactor expansion formula for the cross product of 3 vectors in R⁴.

1a. Directions along A:   v₁ = b−a = (1,0,1) , v₂ = c−a = (1,1,−1).

1b. P(s,t)  =  sv₁ + tv₂ + a  =  (s+t, 1+t, 1+s−t).

1c. The number of independent orthogonal vectors is n−k = 3−2 = 1. We want q = (q₁, q₂, q₃) with q•v₁ = 0 and q•v₂ = 0, giving the linear system:

q₁ + q₃ = 0   ,   q₁ + q₂ − q₃ = 0.

Solving this by elimination gives q₂ = −2 q₁ and q₃ = −q₁, or q = q₁(1,−2,−1) for any q₁ ∈ R. Taking q₁ = 1 gives a specific normal vector q = (1,−2,−1) .

Thus, a defining equation is q•(v−a) = 0, which in coordinates v = (x,y,z) can be simplified to:

x − 2y − z  =  − 3.

We can check that the original points a, b, c all satisfy this equation.

1d. Let r − a = (3,3,4) be the vector from the basepoint a = (0,1,1) on V to the point r = (3,4,5) off of V. The orthogonal distance from V to r can be obtained by projecting r − a on the orthogonal vector q; or simply by dotting r − a with the unit vector ^q⁄_|q|:

(r−a) • ^q⁄_|q|   =   ⁷⁄_√6.

2. The affine line has direction vector v₁ = b−a = (1,0,1), parametrized by c(t) = tv₁ + a = (t,1,t+1).
There are n−k = 3−1 = 2 independent orthogonal vectors satisfying q•v₁ = 0, i.e. q = (q₁, q₂, −q₁) = q₁(1,0,−1) + q₂(0,1,0), giving q₁ = (1,0,−1) and q₂ = (0,1,0). The curve C is the set of points v = (x,y,z) satisfying the two equations q_i•(v−a), i.e.:   x − z = −1  ,  y = 1.

3a. The triple u = (1,1,0), v = (1,0,1), w = (1,1,2):

Notice that we draw the x,y,z axes as a right-handed triple, and v is in the xz-plane in the foreground, whereas w recedes into the picture. This is a left-handed system: model it with your left palm up, index finger horizontal along u, middle finger almost vertical along v, thumb pointing away from you along w. The left-handedness is reflected in the determinant −2, the signed volume of the corresponding parallelopiped, as we will see next time.

We compute the determinant, the signed volume of the parallelepiped spanned by the three vectors, writing u, v, w as the columns of a 3×3 matrix, with one term for each permutation of {1,2,3}:

det(u, v, w)

=

det

1  1 0  1  0 1  1  1 2

=

u1v2w3 − u1v3w2 − u2v1w3 + u2v3w1 + u3v1w2 − u3v2w1

 =  1·0·2 − 1·1·1 − 1·1·2 + 1·1·1 + 0·1·1 − 0·1·1  =  −2.

An easier way to compute this formula is by the cofactor expansion (or minor expansion) along the first column: if we group the above terms by their first factor (u₁, u₂, or u₃), the other factors become 2×2 determinants. That is, we write v⁽ⁱ⁾ for the vector v with its i^th coordinate omitted, and compute:

det(u,v,w)  =  u₁ det(v⁽¹⁾; w⁽¹⁾) − u₂ det(v⁽²⁾; w⁽²⁾) + u₃ det(v⁽³⁾; w⁽³⁾)

 =  1 det(0,1 | 1,2) − 1 det(1,1 | 1,2) + 0 det(1,0 | 1,1)  =  −1 − 1 + 0  =  −2.

The negative value verifies the left-handedness.

3b. The triple u = (1,0,1), v = (1,1,0), w = (1,1,2), with the first two vectors switched from part (a), is right-handed, with the opposite determinant: det(u,v,w) = 2.

3c. The triple u = (1,1,0), v = (1,0,1), w = (2,1,1) has w = u + v, which means the three vectors all lie in the same plane: they are linearly dependent. Thus, the parallelepiped is flat, with volume zero, and we may verify det(u,v,w) = 0. The triple is degenerate, neither right- nor left-handed.

4a. The directions parallel to the plane are u = b − a = (1,0,1) and v = c − a = (1,1,−1), giving the orthogonal direction q = u × v = (−1,2,1).

4b. The equation q • w = q • a becomes (−1,2,1) • (x,y,z) = (−1,2,1) • (0,1,1), or −x + 2y + z = 3. This agrees with the previous HW.

5a. q = (1,1,0) × (1,0,1) = (1,−1,−1)

5b. q • w = (1,−1,−1) • (1,1,2) = −2. This agrees with the previous HW.

6a. The determinant formula defining q = u× is: det(u, w) = q • w. Writing u = (u₁, u₂), w = (x,y), and q = (a,b), this becomes u₁y − u₂x = ax + by, so q = (−u₂, u₁), which is the counterclockwise perpendicular vector to (u₁, u₂). That is, u× = u^⊥.

6b. For three vectors u, v, w ∈ R⁴, their cross product q = u×v×w is the vector orthogonal to all three and with length equal to the volume of the three-dimensional box spanned by them. Its coordinates are the coefficients of x₁, . . . , x₄ in the determinant:

det

u1 u2 u3 u4  v1 v2 v3 v4  w1 w2 w3 w4  x1 x2 x3 x4

.

These can be found using the cofactor expansion along the last column, in which the coefficient of each x_i is given by the determinant of the complementary 3×3 submatrix (with alternating ± signs).