18.1. Math 1920 November 29, ) Solution: In this function P = x 2 y and Q = 0, therefore Q. Converting to polar coordinates, this gives I =

Transcription

1 Homework 1 elected olutions Math 19 November 9, ) olution: In this function P = x y and Q =, therefore Q x P = x. We obtain the following integral: ( Q I = x ydx = x P ) da = x da. onverting to polar coordinates, this gives I = after computing the iterated integral. π 1 r cos θ rdrdθ = π 4, 1) olution: We denote by the path from A to B, and is the region enclosed by and the segment BA. By Green s theorem, ( Q F dr + +BA BA x P ) da Parametrizing AB by 1, t with t from to 1, we get BA ince Q = 4x and P = x, we have Q x P 1 ubstituting this into the first expression above, 1, t, 1 dt = 4 = 4 and hence ( Q x P ) da = 4dA = 4 Area() = 16.. Figure 1: Problem 1 1

2 4) olution: F dr 1 ubstituting the given information gives F dr 1 = 1 It is clear that Area()= 6 4π. Using this above, ( Q x P ) da dxdy = Area(). 1 1 (6 4π) = 1π ) olution: Let R > be sufficiently small so that the circle R is contained in. Let denote the region between R and. We apply Green s Theorem to the region, where the curves and R is are both oriented counterclockwise as in the diagram. This gives F dr R ( Q x P ) da. Now, F x = y x (x + y ) F 1 = y x (x + y ). ince we are given that R π, substituting above we obtain F dr π = da = or π. 7) olution: The vector field (A) does not have spirals nor it is a shear flow. Therefore, the curl appears to be zero. The vector field (B) rotates in the counterclockwise direction, hence we expect the curl to be positive. The vector field () perhaps rotates more strongly clockwise than counterclockwise around the origin, indicating a negative curl (however, this is not completely clear, and concluding that it has a zero curl is also reasonable). Finally, in vector field () the fluid flows straight toward the origin without spiraling. We expect the curl to be zero. ) olution: a) We parametrize the segment from (x 1, y 1 ) to (x, y ) by x = tx + (1 t)x 1, y = ty + (1 t)y 1, t 1. Then, dx = (x x )dt and dy = (y y 1 )dt. Therefore, We obtain the following integral: 1 ydx + xdy = 1 ydx + xdy = (x 1 y x y 1 )dt. 1 (x 1 y x y 1 )dt = 1 (x 1y x y 1 ).

3 b) Let A i = (x i, y i ), i = 1,,..., n, and let be the closed curve determined by the polygon. By the formula for the area enclosed by a simple close curve, the area of the polygon is A = 1 ydx + xdy We use additivity of the line integrals and the result in part (a) to write the integral as follows: ( A = 1 n 1 ) ydx + xdy + ydx + xdy i=1 A ia i+1 A na 1 ( = 1 n 1 ) (x i y i+1 x i+1 y i ) + (x n y 1 x 1 y n ) If we define (x n+1, y n+1 ) = (x 1, y 1 ), we obtain the sum 1 A = 1 n (x i y i+1 x i+1 y i ). i 4) olution: Let be the circle x + y = 9 together with its interior. The divergence of F is so that the flux is F nds = div(f) = 5, div(f)ds = 5 1ds. This integral is five times the area of a circle of radius, so the answer is 5 π = 45π. 4) olution: Using the result we have, using polar coordinates: flux = flux = div(f)da = div(f)da x da = π = 4π. (r cos θ) (r)drdθ 18. 1) olution: We must show that curl(f) d We first compute the line integral around the boundary curve. This curve is the unit circle oriented in the counterclockwise direction, so we parametrize it by Then, γ(t) = (cos t, sin t, ), t π. F(γ(t)) γ (t) = cos t sin t, cos t, sin t sin t, cos t, = cos t sin t + cos t.

4 Then π ( cos t sin t + cos t ) dt = π. Now we compute the flux of the curl through the surface. We parametrize it by Φ(θ, t) = (t cos t, t sin θ, 1 t ), t 1, θ π. Notive that T θ = t sin θ, t cos θ, and T t = cos θ, sin θ, t, so T θ T t = t cos θ, t sin θ, t. ince the normal is supposed to be pointing upward, the z-coordinate of the normal vector must be positive. Therefore, the normal vector is t cos θ, t sin θ, t. The curl in the parameters is: This way, curl(f) = 1,, 1 t cos θ. curl(f) d = π 1 tdtdθ = π. The values of the integrals are equal, as stated in tokes Theorem. 6) olution: curl(f) is xy curl(f) = y + 1 z, y + 1, 1. We compute the flux of the curl through the surface by using tokes Theorem and computing the line integral around the boundary circle. The oriented boundary of this surface is the triangle at height z =, oriented clockwise (when viewed from above). By tokes Theorem, the flux of the curl is equal to the line integral of F around the oriented boundary. The restriction of F to the boundary, where z = is F = x + y,, x y + 1. Next, parametrize the three sides of this triangle for t 1: Thus ds on the three sides is 1 t,,,, t,, t, 1 t,. 1,, dt,, 1, dt, 1, 1, dt, and the dot products are (the z-components of F is not relevant because ds has zero z-component): F ds = 1 t,, 1,, dt = (t 1)dt F ds = t,,, 1, dt = F ds = 1,, 1, 1, dt = dt Therefore, the line integral around the oriented boundary is equal to 4

5 1 (t 1)dt dt = 1, and thus the flux of curl(f) through the surface is 1. 15) olution: (a) The induced orientation is defined so that as the normal vector travels along the boundary curve, the surface lies to its left. Therefore, the boundary circles on top and bottom have opposite orientations, which are shown in the figure below. (b) We first compute the integral around the boundary circles using the following parametrizations: Observe that: 1 : γ 1 (t) = ( cos t, sin t, 6), t from π to 1 : γ 1 (t) = ( cos t, sin t, 1), t from to π F(γ 1 (t)) γ 1(t) = 7 sin t,, sin t, cos t, = 144 sin t. F(γ (t)) γ (t) = sin t,, sin t, cos t, = 4 sin t. The line integral is thus F dr + 1 The curl is ( 144 sin t)dt + π π ( 4 sin t)dt = 14π. We parametrize by curl(f) =, yz, z. Φ(θ, z) = ( cos θ, sin θ, z), θ π, 1 z 6. The outward pointing normal is cos θ, sin θ,, hence curl(f)(φ(θ, z)) N =, 4z sin θ, z cos θ, sin θ, = 8z sin θ We obtain the following integral: curl(f) d = 6 π 1 8z sin θdθdz = 14π. The line integral and the flux have the same value. This verifies tokes Theorem. Figure : Problem 15a 5

6 18) olution: We first compute the surface integral directly. The spherical cap is parametrized by Φ(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ), θ π, φ π. The outward pointing normal is sin φ(cos θ sin φ, sin θ sin φ, cos φ), hence We obtain the following integral: F(Φ(θ, φ)) N =, cos φ, 1 N = sin θ sin φ cos φ + sin φ cos φ. F d = π π ( sin θ sin φ cos φ + sin φ cos φ)dφdθ = π 4. We now evaluate the flux using tokes Theorem. A straightforward computation shows that F = curl(a), where A =, x, xz. By tokes, F d = curl(a) d = A dr. To compute the line integral, we notice that the boundary curve is the circle x + y = 4 z = 1. We parametrize by ( γ(t) = cos t, ) sin t, 1, t π. in the plane Hence, A(γ(t)) γ (t) =, cos t, 4 cos t sin t, cos t, = 4 cos t. We obtain the following line integral: which implies A dr = π 4 cos tdt = π 4, F d = π 4. This agrees with the other computation of the flux, as expected. Figure : Problem 18 6

7 ) olution: ince we are interested in F ds, we can also consider curl(f) d, by tokes Theorem. The curl is 4y,, 1 y anf the normal to the plane is a, b, c. They are orthogonal if which means 4y,, 1 y a, b, c =, (4a c)y + (c a) =. We conclude that c = a and b arbitrary solves the problem. In other words, the plane ax+by+az = does the job. 7) olution: Let be the unit circle centered at the origin in the xy-plane. ince F has a vector potential (say, A), we have by tokes Theorem: F d = curl(a) d = A dr = 5. 7