HOMEWORK 8 SOLUTIONS

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1 HOMEWOK 8 OLUTION. Let and φ = xdy dz + ydz dx + zdx dy. let be the disk at height given by: : x + y, z =, let X be the region in 3 bounded by the cone and the disk. We orient X via dx dy dz, then by definition the boundary orientation of agrees with the orientation by the outward-pointing normal vector. By tokes Theorem we have: X dφ = We claim that X φ = φ =. φ + φ. We check this on each of the three terms of φ. The integral of the third term over is zero since the function in that term is z and is contained inside the plane z =. The other two terms integrate to zero over because the value of z is constant over and those two terms contain a dz. We compute: dφ = 3dx dy dz. ince X is oriented by dx dy dz, the integral of dφ over X will be equal to 3 times the volume of X. To compute this we use the well-known formula for the volume of the cone V = 3 π h. In our case the volume is 3 πa3. Putting everything together, we find φ =. X dφ = 3Vol(X) = πa 3. (a) The derivative of the flux form is the mass form of the divergence. Note that r = x x xn x n = n hence dφ F = M n = ndx... dx n.

2 (b) Apply tokes to the form ω = Φ F and X = B n () the ball of radius in n whose boundary is n. We have Φ F = n dφ F = n B n dx... dx n = nvol n (B n ). B n Now, for the left hand side, we find the flux by computing F n = x... x... = hence This yields x n Φ F = n F n d = n d = vol n ( n ). n 3. We have seen in class that for any form ω. Now specialize to and use that ubstituting, we obtain which is what we needed. M x n vol n ( n ) = nvol n (B n ). M dω = ω = α β dω = dα β + ( ) k α dβ. dα β + ( ) k 4. By Green s theorem (x y + y 3 y)dx + (3x + y x)dy = where is the region bounded by. Note that = M α dβ = 4 x y dxdy, (3 + y ) (x + 3y ) dxdy the integrand is positive everywhere inside the disk of radius, zero on the circle of radius, and negative outside the disk. Imagine the region to be a membrane expanding starting from the origin. As long as stays inside the disk of radius, the integral gets larger and larger since the integrand is positive. As soon as leaves the disk, the integrand becomes negative, and the integral will start decreasing. o to maximize the integral, should be the oriented boundary of the disk, i.e. the circle of radius oriented in the counter-clockwise direction.

3 5. The maximum value is 4 x y = (i) We find o π (4 r )r dr dθ = π [4r r4 4 ] r= = 8π. F = x (xy x4 y) + y (4x + 4x3 y y ) = y 8x 3 y + 8x 3 y y =. for any region. Let F da = and be two curves in the first quadrant from the line x = to the line x =. onsider in addition: 3 is the segment of the line x = from the endpoint of to the endpoint of, 4 is the segment of the line x = joining the endpoint of to the endpoint of. The four curves, 3,, and 4 together bound a region. The flux through 3 and 4 is found to be zero: xy x 4 y dy = y y dy = 3 3 and (xy x 4 y)( dy) = 4 dy =. 4 Now we use Green s theorem: = F da = Φ F + Φ F + 3 Φ F + = Φ F Φ F, so Φ F = Φ F. Therefore the flux through any two such curves is the same. 4 Φ F emark: and might intersect, making the union of several regions with different orientations, but this is okay because the area integral F da is still zero. 6. (ii) To find the value of the flux, we choose the path along the x axis, on which y =. [ ] (4x + 4x 3 y y ) dx = 4x dx = x =.

4 (i) We have F = x ( ) x x + y + ( ) y y x + y = y x x + y + x y x + y =. (ii) Parameterize the circle by (x, y) = (a cos t, a sin t), t π, M dy N dx = π = (dx, dy) = ( a sin t dt, a cos t dt). a cos t a cos t + a sin t a cos t a sin t a cos t + a sin ( a sin t) dt t π cos t + sin t dt = π dt = π. emark: Note that the flux is nonzero even though the divergence F is zero. Green s theorem doesn t apply in this case because F isn t defined everywhere inside the disk. (iii) For two closed curves and enclosing the origin, the curves and enclose a region which does not include the origin. o Green s theorem applies to : Φ F Φ F = F da = da =. o Φ F = and the flux is the same for all such curves. Φ F emark: and might intersect, making the union of several regions with different orientations, but this is okay because the area integral F da is still zero. 7. (i) We compute the integral by Green s theorem. First, dw F = d((f(x) + y )dx + (g(y) + xy)dy) = ydx dy. We have W F = dw F = ydx dy. This integral equals zero by symmetry, since the negative values of y cancel the positive values of y. (ii) This is not true. We have seen in class that the integral of W F is path independent if and only if dw F =. However, dw F = ydx dy. 8. Let M = ax y 3y xy + and N = x 3 y bx x y

5 be the two components of the field. Then M y = ax 3 xy, N x = 3x b xy. Then, since the two dimensional curl is zero, M y = N x, which implies that a = b = 3. To find a potential, note that Then A potential is f x = 3x y 3y xy + = f = x 3 y 3xy x y + x + g(y). f y = x 3 3x x y + g (y) = x 3 y 3x x y = g (y) = y = g(y) = y. 9. First note that r = (x + y ). The curl is f = x 3 y 3xy x y + x y. x (x + y ) n y y (x + y ) n n x = (x + y ) n (xy yx) =. Let us find a potential f. ince f x = x(x + y ) n we integrate with respect to x to find Next, we compute Thus f(x, y) = (x + y ) n (x + y ) n+ x dx = n + + g(y). f y = (x + y ) n y + g (y) = (x + y ) n y g (y) = = g(y) =. f(x, y) = (x + y ) n+ n + + = rn+ n + +. A different discussion is necessary when n =, since in this case the integration above has a different answer: f = ln r +.. We have We need We compute Thus F = (axy + z 3 )i + (x + byz)j + (y + cxz + )k. F =. F = y by, 3z cz, x ax. a = b =, c = 3.

6 To find a potential f we need to have F = f or f x = xy + z 3, f y = x + yz, f z = y + 3xz +. The first equation gives f = x y + xz 3 + g(y, z). Thus differentiating with respect to y and using the second equation f y = x + g y = x + yz = y y = yz = g = y z + h(z). We have f = x y + xz 3 + y z + h(z). ifferentiating with respect to z and using the value of f z we find f z = 3xz + y + h (z) = h (z) = = h(z) = z +. Thus f = x y + xz 3 + y z + z +.. onsider the original surface = {z = e x y }. We close the surface by adding a disk in the z = plane. That is, we consider the surface given by x + y z =. Let be the region enclosed by and. We need to orient by the normal pointing down. We have dφ F = M divf = ( + )dx dy dz =. By the ivergence Theorem Φ F + Φ F = dφ F =. To find the value of the flux over note that the normal vector is n =,, so F n = x, y, z,, = z =. The flux over is therefore equal to, so. Let Φ F =.

7 be the surface of the paraboloid : z = x + y, x + y ; oriented by the downward-pointing normal is the flat disc at height : : z =, x + y oriented by the upward-pointing normal is the region between and so that the boundary = + when is oriented standardly by dx dy dz. We need to compute both sides of the divergence theorem: Φ F = M F. For the left hand side, we parametrize and : ( ) γ, γ : [, ] [, π] 3 r, γ = r cos θ ( ) r sin θ r, γ θ = r cos θ r sin θ. θ We have: γ = cos θ r sin θ sin θ r cos θ, γ = cos θ r sin θ sin θ r cos θ. r One can check that γ is orientation-reversing and γ is orientation-preserving by computing r the cross products of columns γ γ and γ γ and checking against the normal vectors orienting and. Now, we find the pullbacks: Φ F = ydy dz + (x + )dx dz + z 3 dx dy, γ (Φ F ) = r 3 sin θ cos θ + (r cos θ + )r sin θ + r 7 = (r sin θ + r 7 )drdθ, γ (Φ F ) = (r cos θ + r sin θ) = rdrdθ. We integrate and change signs to account for the orientations π π [ Φ F = (r sin θ + r 7 )drdθ = 3 r3 sin θ + 8 r8 π ( = 3 sin θ + ) [ dθ = 8 3 cos θ + θ ] π 8 imilarly, Φ F = π rdrdθ = π. θ= = π 4. ] dθ = r=

8 ollecting the answers we have Φ F = Φ F + Φ F = 3π 4. For the right hand side, we have dφ F = M div F = 3z dx dy dz. Parametrizing by cylindrical coordinates gives: π M div F = 3z rdzdrdθ = π r [ r = π (r r 7 )dr = π r8 8 Thus the ivergence Theorem is verified. ] r= [ z 3 r ] z=r dr = = 3π We check Φ F = W F. We evaluate the left hand side: F = (y y, x x, ) = (y, x, ) We need to integrate Φ F = y dy dz + xdx dz. The surface is a portion of the graph of z = g(x, y) = 5 5 (x + y ) x + y. We parametrize via We have The pullback Thus ([ x γ = y]) x y 5 5 (x + y ) γ = 5 x 5 y, γ Φ F = y 5 x x 5 y =. Φ F =.

9 For the right hand side, is the tip of a paraboloid that opens downward. Its boundary is the circle of radius in the xy plane where z =. Thus W F = xz dx + yz dy + (x + y ) dz = along z =. Thus the line integral along is W F =. 4. We need to check W F = Φ F. For the right hand side, begin by computing the curl: i j k F = x y z = (4 + )i + ( + 3)j + ( )k = 5i + j k y z x + y z 4y 3x The normal vector equals n = x, y, z, so that F n = 5 x + y z. We need to integrate this quantity over the lower half sphere. By symmetry, the integrals of x and y are. Thus the surface integral equals zd = π π cos φ 4 sin φdφdθ = 8π π/ For the line integral on the left hand side, we have z =, so We need to evaluate W F = ydx + (x + y )dy. y dx + (x + y )dy, π π/ sin φ cos φdφ = 4π. where is the circle of radius in the xy plane oriented clockwise, enclosing a disk. We can parametrize, or alternatively, we could use Green s Theorem to evaluate the line integral in the plane, including a sign in front of the integral to account for the negative orientation. Then, W F = tokes Theorem is verified. ( x (x + y ) ) y (y) da = da = area () = 4π.

10 5. (i) We compute the curl F = F 3 3 F 3 F F 3 = z. F F y (ii) We have F d = F nd, where n is the outward-pointing unit normal vector. ince is contained in the sphere of radius a, we have that the normal vector at the point (x, y, z) on is given by: n = x/a y/a. z/a Then by part (a) we have F n = on all of. Thus the flux of F through is zero. (iii) This is immediate from tokes Theorem and (ii): W F = dw F = Φ F =. 6. We need to check that W F = For the left hand side, note that is given by: Φ F. : x + y = 3, z =. We parametrize as follows: γ : [, π] 3 3 cos θ, γ(θ) = 3 sin θ. We check that γ is orientation-preserving, by verifying the right hand rule. We have: W F = yzdx + xzdy + z 3 dz, γ (W F ) = ( 3 sin θ + 3 cos θ) dθ = 3 cos θdθ, π [ ] π W F = 3 cos θdθ = 3 sin θ =. θ=

11 For the right hand side, we compute F = x y, dw F = Φ F = xdy dz ydx dz. We parametrize using spherical coordinates. Note that largest value of φ occuring for (which occurs on ) is equal to π/3. The normal vector is We compute n = x, y, z. F n = x, y, x, y, z = ( x + y ) = ( cos θ sin φ + sin θ sin φ) = cos θ sin φ. In spherical coordinates Thus the flux equals d = 4 sin φ dφ dθ. π π/3 F n d = 8 sin 3 φ cos θdφdθ = ( π ) ( π/3 ) = cos θdθ 8 sin 3 φdφ = since the first of the two integrals is zero as calculated above. Therefore tokes Theorem is verified.

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