Ma 1c Practical - Solutions to Homework Set 7

Transcription

1 Ma 1c Practical - olutions to omework et 7 All exercises are from the Vector Calculus text, Marsden and Tromba (Fifth Edition) Exercise Find the area of the portion of the unit sphere that is cut out by the cone z x + y. olution. The intersection of the unit sphere and the cone z = x + y is found by solving the equations x + y + z = 1 and x + y z = (with z ), which is easily done by subtracting these two equations. This gives the circle described by z = 1/ and x + y = 1/, as in Figure 1. We are to find the area of the surface above this circle. z A B z = 1/ O y x Figure 1. Find the area of the ice cream part of this surface. Notice that the triangle AOB has two sides of length 1/, and hypotenuse of length 1, so the vertex angle AOB is π/4. Using this geometry and spherical coordinates, we find that a parametrization is x = sin φ cos θ y = sin φ sin θ z = cos φ, for θ π, φ π 4. We find that π 4 Area = sin φ dφ dθ ( ) = 1 π = ( )π. 1

2 Exercise Evaluate xyz d where is the triangle with vertices (1,, ), (,, ) and (, 1, 1). olution. The triangle is contained in a plane whose equation is of the form ax + by + cz + d =. ince (1,, ) lies on it, a + d =, so a = d. ince (,, ) is on it, b = 1 d. ince (, 1, 1) is on it, b + c = d, so c = d + 1 d = 1 d. Letting d =, we get x + y + z = i.e., the equation of the plane is given by x + y + z =. ee Figure A normal vector is obtained from the coefficients as (, 1, 1), so a unit normal is n = 1 (i + j + k). The domain in the xy plane is the triangle with vertices (1, ), (, ) and (, 1), as in Figure. y (,) z (,1,1) (,1) (1,) x (1,,1) (,,) x y Figure. The domain in the xy-plane for plane in Exercise regarded as a graph: z = x y. Now and so f d = d = = dx dy n k = dx dy, xy( x y) dx dy 1 (1 x) 1 x [(x x )y xy ]dy dx

3 3 Carrying out the y-integration gives f d = 1 ) ((x x ) y xy3 (1 x) dx 3 1 x = 1 [ ( ) [(1 x)] [1 x] x(1 x) x ] 3 ([(1 x)]3 (1 x) 3 ) dx = 1 3 x(1 x)3 dx = 1 where the last steps were done using integration by parts (1 x)4 dx = 3, Exercise Let be the closed surface that consists of the hemisphere x + y +z = 1, z, and its base x +y 1, z =. Let E be the electric field defined by E(x, y, z) = xi + yj + zk. Find the electric flux across. olution. Write = where is the upper hemisphere and is the disk. ence E d = E d + E d. (i) Let xi + yj + zk be the unit normal n pointing outward from. Then E d = E n d = (x, y, z) (x, y, z)d = (x + y + z )d = d = 4π. (ii) The unit normal is k and z = on. ence, E d = E n d = (x, y, z) (,, 1)d =. Therefore, E d = 4π. Exercise Let the velocity field of a fluid be given by v = i + xj + zk in meters/second. ow many cubic meters of fluid per second are crossing the surface x + y + z = 1, z? (istances are in meters.) olution. ere, v d = v n d and n = xi + yj + zk, so v n = x + xy + z. By symmetry, the integrals of x and of xy vanish. Thus, the flux is v d = z d. Using spherical coordinates, z = cos φ, so we get π/ φ= θ= cos φ sin φ dθ dφ = π cos3 φ 3 π/ = π 3.

4 4 Exercise 8.1.3(d). Verify Green s theorem for the disk with center (, ) and radius R for P = y, Q = x. olution. Green s theorem P dx + Qdy = becomes y dx + x dy = ( Q x P ) dx dy y (1 )dx dy = dx dy The right side is πr while the left side is, since x = R cos θ and y = R sin θ, (R sin θ)( R sin θ)dθ + (R cos θ)(r cos θ)dθ = R sin θdθ + R cos θdθ. Using the fact that sin θ and cos θ have averages 1, namely 1 π sin θdθ = 1 (this is one way of remembering the formula for the integrals of sin θ and cos θ on [, π] and [, π]), we get R π + R π = πr. Thus, Green s theorem checks. Exercise Find the surface integral ( F) d, where is the ellipsoid x + y + z = 1 and F = (sin xy)i + e x j yzk. olution. Notice that the ellipsoid is a closed surface and has no boundary. Therefore, by tokes theorem, ( F) d = F d =. Note: The same conclusion also follows from the divergence theorem since div curl F =. Exercise Let F = x i + (xy + x)j + zk. Let C be the circle x + y = 1 in the plane z = oriented counterclockwise and the disk x + y 1 oriented with the normal vector k. etermine: (a) The integral of F over. (b) The circulation of F around C. (c) Find the integral of F over. Verify tokes theorem directly in this case. olution. (a) Notice that F = (x, xy + x, ) on. ence F d = (x, xy + x, ) (,, 1) d =.

5 (b) (c) Let c(t) = (cos t, sin t, ) be the parameterization of C. Then F d = (cos t, cos t sin t + cos t, ) ( sin t, cos t, )dt C = (cos t sin t + cos t)dt = π. Routine computation shows that F = (,, y + 1). ence 1 ( F) d = (,, r sin θ + 1) (,, 1)r dθ dr = 1 (r sin θ + 1)r dθ dr = π. Combining the results in (b) and (c), tokes theorem is verified. Exercise etermine which of the following vector fields F in the plane is the gradient of a scalar function f. If such an f exists, find it. (a) F(x, y) = (cos xy xy sin xy)i (x sin xy)j (b) F(x, y) = (x x y + 1)i + (y x y + 1)j (c) F(x, y) = (x cos y + cos y)i (x sin y + x sin y)j. olution. In this problem, we apply the cross-derivative test. problem (a), 5 For example, for F 1 y F x = (x sin xy x sin xy x y cos xy) ( x sin xy x y cos xy) =, so F is indeed the gradient of some function on the plane. To find such a function, we seek f satisfying f y = F = x sin xy, for example, f(x, y) = x cos xy. Note that f is unique only up to an additive that could be a function of x. owever, we don t need to add it in this case as this function is checked to have gradient the given vector field. Part (b) and (c) proceed similarly. One sees that (b) is not a gradient field, while (c) is a gradient. For part (c), f(x, y) = x cos y + x cos y is a function whose gradient is the given field. Exercise Evaluate the surface integral F n d, where F(x, y, z) = i + j + z(x + y ) k and is the surface of the cylinder x + y 1, z 1, including the sides and both lids. olution. Use Gauss divergence theorem in space: F n d = (div F)dx dy dz ere, div F = x (1) + y (1) + z z(x + y ) = (x + y ). The region W is a cylinder, so it is the easiest to evaluate the integral in cylindrical coordinates: W

6 1 1 r (r ) dr dθ dz = π = π 3. Exercise Fix k vectors v 1,..., v k in space and numbers ( charges ) q 1,..., q k. efine k q i φ(x, y, z) = 4π r v i, where r = (x, y, z). how that for a closed surface and e = φ, e d = Q, i=1 where Q = q q k is the total charge inside. charges are on. Assume that none of the olution. urround each charge at vector v i by a small ball B i in such a way that the B i are mutually disjoint and do not intersect. Assume that B 1,..., B n, (where n k) are those balls contained within. Then since div e =, and as in Theorem 1, n e d = e d i=1 B i where B i is given the outward orientation. But again, as in Theorem 1, e d = q i. B i Thus, n e d = q i = Q, the total charge inside. i=1