Math 31CH - Spring Final Exam

Transcription

1 Math 3H - Spring 24 - Final Exam Problem. The parabolic cylinder y = x 2 (aligned along the z-axis) is cut by the planes y =, z = and z = y. Find the volume of the solid thus obtained. Solution:We calculate = x 2 y 2 ( x2 ) 2 dx = dz dy dx = x 2 ( x 2 ) 2 dx = y dy dx = 2 y2 z= x2 y= dy ( 2x 2 + x 4 ) dx = = 8 5.

2 Problem 2. Find the volume of the region in the first octant bounded by the hyperbolic cylinders yz 4, xz 9, 4 xy 9. Solution:Use the change of coordinates It is easy to see that Since uvw = (xyz) 2, this rewrites as This gives The volume becomes dx dy dz = u = yz, v = xz, w = xy. (u, v, w) (x, y, z) = (u, v, w) (x, y, z) = 2 uvw = z y z x y x = 2xyz. (x, y, z) (u, v, w) =. 2 uvw dx dy dz = du dv dw. 2 uvw 2 u /2 v /2 w /2 du dv dw = 2 2u 2 u=4 u= 2v 2 u=9 v= 2w 2 w=9 w=4 = 8.

3 Problem 3. onsider the field F = axyz 3 + 4xy, bx 2 z 3 + cx 2 + 2z 2, 3x 2 yz 2 + 4yz +. (i) For what values of the constants a, b, c, is F a gradient field? (ii) Find a potential function for the field F. (iii) Find the integral W F where R 3 is the curve parametrized by Solution: γ : [, ] R 3, γ(t) = (e t sin(πt 4 ), t 4 te t3, t 5 ). (i) We need F =. alculating the determinant i j k F = x y z axyz 3 + 4xy bx 2 z 3 + cx 2 + 2z 2 3x 2 yz 2 + 4yz + we obtain (ii) We need = (3 3b)x 2 z 2, ( 6 + 3a)xyz 2, (2b a)xz 3 + (2c 4)x = a = 2, b =, c = 2. f x = 2xyz 3 + 4xy = f = x 2 yz 3 + 2x 2 y + g(y, z) f y = x 2 z 3 + 2x 2 + 2z 2 = g y = 2z 2 = g = 2yz 2 + h(z) = f = x 2 yz 3 + 2x 2 y + 2yz 2 + h(z) Thus is a potential. f z = 3x 2 yz 2 + 4yz + = h z = = h = z. f = x 2 yz 3 + 2x 2 y + 2yz 2 + z (iii) By the fundamental theorem of calculus for line integrals, W F = f(γ()) f(γ()) = f((,, )) f((,, )) =.

4 Problem 4. Fix a, h >. onsider the field F = x 2, xy, 2xz. Let D R 3 be the half solid cylinder contained between: the cylindrical surface x 2 + y 2 = a 2 and the yz-plane (laterally); the plane z = on the bottom, and the plane z = h on top. (i) Find the outward flux of the field over the boundary of D. Make sure you integrate over each of the four pieces of the boundary. (ii) alculate the divergence of the field F. Verify that the divergence theorem holds true for the field F and the closed cylinder above. That is, write down the divergence theorem, compute both sides, and check that they are equal. Solution:We assume that the half cylinder has x. (The other half is similar.) (i) We calculate the flux over the four pieces of the boundary: bottom disk: the normal vector is n =,, and F n = 2xz = since z = over the disk. Thus the flux is zero. top disk: the normal vector is n =,, and F n = 2xz = 2xh since z = h. The 2xh dx dy = 2h flux is computed in polar coordinates to be π/2 a π/2 r cos θ(r dr) dθ = 2h a π/2 r 2 dr cos θ dθ = 2h a3 π/2 3 2 = h 4a3 3. the yz-plane: the normal vector is n =,, and F n = x 2 = since x = over this plane. the surface of the cylinder: the normal vector is n = a x, y, so F n = a x2, xy, 2xz x, y, = a (x3 + xy 2 ) = a x(x2 + y 2 ) = ax. We integrate in cylindrical coordinates π/2 h h ax ds = a(a cos θ) adzdθ = a 3 π/2 Putting things together, the total flux is 2a 3 h 4a3 h 3 = 2a3 h 3. π/2 dz cos θ dθ = a 3 h 2. π/2 (ii) We have F = x, y, z x 2, xy, 2xz = 2x + x 2x = x.

5 We integrate in cylindrical coordinates π/2 a h F dv = x dv = (r cos θ) dz(rdr)dθ = D D π/2 = h a This is exactly the answer we found in (i). h dz a π/2 r 2 dr cos θdθ π/2

6 Problem 5. onsider the surface S R 4 given by the equations oriented by the form in a neighborhood of the point p = (,,, ). p = (,,, ). x 2 yz + zw =, xw 2 + x 3 + zw = 2 Ω = dy dw Write down a positive basis of S at the point Solution:We compute the tangent space of the surface at p = (,,, ). The matrix of partial derivatives equals [ ] 2x z y + w z w 2 + 3x 2. w 2xw + z At the given point p, this matrix equals [ ] The tangent space is the null space of this matrix. Solving for the null space, we see that any vector (x, y, z, w) in the tangent space satisfies the system Thus 2x =, 4x + z + 2w =. x =, z = 2w. Thus vectors in the tagent space have the form x y z = y + w 2. w For the basis we have so this basis is positive., 2 Ω, 2 = = >

7 Problem 6. onsider the surface M R 3 given by x 4 + y 4 + z 4 = and let F = x 3 z, y 3 z, z. (i) alculate F, and show that for any piece X of the surface M we have Φ F =. (ii) Show that for any closed curve on the surface M, we have W F =. Solution: (i) We have i j k F = x y z x 3 z y 3 z z = y3, x 3,. The normal vector is n = 4x 3, 4y 3, 4z 3 so F n = y 3, x 3, 4x 3, 4y 3, 4z 3 =. Thus Φ F = M X M F n ds =. (ii) Assume that bounds a piece X of the surface M. By Stokes, we have W F = Φ F = using part (i). X

8 Problem 7. Let: onsider the field F = 2xz, x, y 2. be the circle at the intersection of the paraboloid z = x 2 + y 2 and the sphere x 2 + y 2 + z 2 = 2; P be the piece of the paraboloid below ; S be the spherical cap above. Orient counterclockwise, and orient P and S compatibly. (i) Find the curl F and write down the flux form Φ F. (ii) Evaluate the flux of the curl of F Φ F P over the piece P of the paraboloid. (iii) Evaluate the flux of the curl of F over the spherical cap Φ F. S (iv) The answers in (ii) and (iii) should agree. Explain why you expect this to happen. Solution: (i) We have The flux form is i j k F = x y z = 2y, 2x,. 2xz x y 2 Φ F = 2y dy dz + 2xdx dz + dx dy. (ii) We parametrize the paraboloid by γ(x, y) = (x, y, x 2 + y 2 ) of the paraboloid P defined for x 2 + y 2. We find Dγ =. 2x 2y

9 We check this parametrization respects the orientation. Indeed, P is oriented by the upward pointing normal. Since D γ D 2 γ has the last component +, this vector is upward pointing as well, hence the parametrization respects the orientation. We pullback the integrand under γ, thus obtaining γ (2y dy dz + 2x dx dz + dx dy) = {2ydy dz (D γ, D 2 γ) + 2x dx dz (D γ, D 2 γ) + dx dy (D γ, D 2 γ)} dx dy = {(2y)( 2x) + (2x)(2y) + } dx dy = dx dy. We now integrate the pullback over the domain of the parametrization dx dy = Area = π. x 2 +y 2 (iii) The normal vector of the spherical cap is n = 2 x, y, z, oriented upward compatibly with. Then F n = 2y, 2x, 2 x, y, z = z 2 = cos φ, using spherical coordinates. We have ds = 2 sin φ dφdθ and We integrate F n ds = S 2π π/4 φ π 4, θ 2π. 2 sin φ cos φ dφdθ = 2π dθ π/4 2 sin φ cos φ dφ = 2π sin 2 φ π/4 φ= = π. (iv) The answers in (ii) and (iii) agree because of Stokes theorem. By Stokes, both flux integrals are in fact equal to the work integral over the boundary curve W F. Another explanation is possible via the divergence theorem applied to the region D between P and S: since S Φ F Φ F = dφ F = M ( F) = P D D ( F) =.

10 Problem 8. onsider the manifold M 2 consisting of pairs (z, w) 2 such that w = z + e z. What is the surface area of M contained over the region where z = x + iy satisfies x, y 2π. Solution:We have w = (x + iy) + e x+iy = (x + iy) + e x (cos y + i sin y) = (x + e x cos y) + i(y + e x sin y). We parametrize Thus From here, we find Thus Dγ T Dγ = = ([ x x γ : [, ] [, 2π] M, γ = y y]) e x cos y + x. e x sin y + y Dγ = e x cos y + e x sin y. e x sin y e x cos y + [ e x cos y + e x ] sin y e x sin y e x cos y + e x cos y + e x sin y e x sin y e x cos y + ]. [ + (e x cos y + ) 2 + e 2x sin 2 y + (e x cos y + ) 2 + e 2x sin 2 y det DγT Dγ = + (e x cos y + ) 2 + e 2x sin 2 y = 2 + e 2x (cos 2 y + sin 2 y) + 2e x cos y We now integrate 2π 2 + e 2x + 2e x cos y dy dx = = 2 + e 2x + 2e x cos y. 2π(2 + e 2x ) dx = 2π(2x + 2 e2x x= x=) = π(e 2 + 3).

11 Problem 9. Fix a,..., a n > be positive numbers. onsider the ellipsoid M R n given by x 2 a 2 and oriented by the outward normal. the (n )-form ω in R n given by x2 n a 2 n ω = x dx 2 dx 3... dx n x 2 dx dx 3... dx n ( ) n x n dx... dx n. (i) ompute dω. (ii) Express the volume of the solid ellipsoid x 2 a x2 n a 2 n in terms of the volume β n of the unit ball in R n and the constants a,..., a n. (iii) Determine M ω. = Solution: (i) We have ω is the flux form of the field x F = x x n Thus F = n. We find dω = dφ F = M F = ndx dx 2... dx n. (ii) With the change of variables x i = a i y i we have dx... dx n = a... a n dy... dy n. In the new coordinates, the solid ellipsoid E becomes y yn 2 which is the ball B n of radius. Thus vol n (E) = dx... dx n = a... a n dy... dy n = a... a n vol n (B n ) = a... a n β n. E B n (iii) The boundary of the solid ellipsoid E is the manifold M. From the generalized Stokes theorem, we find ω = M E dω = E ndx... dx n = nvol n (E) = na... a n β n.

12 Problem. It is known that the form ω = ydx xdy x 2 + y 2 is closed wherever defined i.e. dω =. Determine all possible values of the integral ω where is an oriented curve (without self-intersections) joining (, ) to (2, ), not passing through the origin. (i) First determine the integral along a straight line segment L joining (, ) to (2, ). (ii) Next, consider an arbitrary curve and compare the answer with the answer you found in (i) using Green s theorem for a suitable region. Solution: (i) We parametrize the line segment as for t. Then ( ) ydx xdy γ x 2 + y 2 = Integrating, we obtain x = + t, y = t ( t) d( + t) ( + t)d( t) ( + t) 2 + t 2 = 2 dt (t + /2) 2 + /4 = 3/2 /2 dt 2t 2 + 2t + = 2 2 du u 2 + (/2) 2 = arctan(2u) u=3/2 u=/2 dt (t + /2) 2 + /4. = arctan(3) arctan() = arctan(3) π 4. Note that it may be possible that different ways of parametrizing the segment yield the answer in a seemingly different form: arctan(/2) is an example. This is equivalent to the answer found above via trigonometric manipulations. (ii) We let: is a curve joining (, ) to (2, ); R is the region between and L; we give L the opposite orientation as in part (i), so that the boundary of R consists in the curves and L compatibly oriented. Two cases will be distinguished:

13 If the origin does not belong to the region R, we can apply Green s theorem to R. We have dω = R ω ω = = L ω = L ω = arctan(3) π 4. If the origin belongs to the region R, we enclose the origin by a small circle a of radius a. We let R a be the region between the circle a, the line L and the curve we give L the opposite orientation as in part (i) and give a the clockwise orientation. Then the boundary of R a consists in, L, a with the given orientations being the correct orientations. By Green s theorem applied to the region R a, we have dω = ω ω + ω = = ω = ω ω. R a L a L a Now the integral over a is found by polar parametrization x = a cos t, y = a sin t and changing the sign of the integral to account for orientations 2π a(sin t)d(a cos t) (a cos t)d(a sin t) 2π ω = a a 2 (cos t) 2 + a 2 (sin t) 2 = dt = 2π. Thus ω = arctan(3) π 4 2π = arctan(3) 9π 4.