MTH U341 urface Integrals, tokes theorem, the divergence theorem To be turned in Wed., Dec. 1. 1. Let be the sphere of radius a, x 2 + y 2 + z 2 a 2. a. Use spherical coordinates (with ρ a) to parametrize. r(φ, θ) <a sin φ cos θ, a sin φ sin θ, a cos φ>. b. Use the parametrization from (a) to compute the area of as d. Be sure to substitute for d! ecall that d r φ r θ. so r φ <a cos φ cos θ, a cos φ sin θ, a sin φ> r θ < a sin φ sin θ, a sin φ cos θ, >, i j k r φ r θ a cos φ cos θ a cos φ sin θ a sin φ a sin φ sin θ a sin φ cos θ a 2 sin 2 φ cos θi + a 2 sin 2 φ sin θj + a 2 cos φ sin φ k a 2 sin φ<sin φ cos θ, sin φ sin θ, cos φ>. r φ r θ a 2 sin φ<sin φ cos θ, sin φ sin θ, cos φ> a 2 sin φ <sin φ cos θ, sin φ sin θ, cos φ> a 2 sin φ sin 2 φ cos 2 θ + sin 2 φ sin 2 θ + cos 2 φ a 2 sin φ sin 2 φ(cos 2 θ + sin 2 θ) + cos 2 φ a 2 sin φ sin 2 φ + cos 2 φ a 2 sin φ. We can now compute the integral: d 2π π 2π 2π 2π a 2 sin φ dφ dθ a 2 cos φ π, dθ a 2 (cos(π) cos()), dθ 2a 2 dθ 4πa 2. 2. Let be the surface z x 2 + 3y 2, x 2, y 3. a. Parametrize using the standard parametrization of a graph. r(x, y) <x, y, x 2 + 3y 2 > 1
that is, x x, y y, z x 2 + 3y 2. b. Let r(x, y) be the parametrization from (a). Compute the unit normal explicitly. What is n(, )? n(x, y) 1 r x r y r x r y r x <1,, 2x>, r y <, 1, 6y> and i j k r x r y 1 2x 1 6y < 2x, 6y, 1>. and < 2x, 6y, 1> 4x 2 + 36y 2 + 1 n(x, y) 1 < 2x, 6y, 1> 4x2 + 36y 2 + 1 n(, ) <,, 1>. c. Let F be the vector field F (x, y, z) yi + x 2 j + (4z + y 2 x 2 ) k. Compute F d, where is oriented using the unit normal from (b). F d where is the rectangle x 2, y 3. Also F (r(x, y)) r x r y da F (r(x, y)) <y, x 2, 4(x 2 + 3y 2 ) + y 2 x 2 > <y, x 2, 3x 2 + 13y 2 >, so from (2b) we get F (r(x, y)) r x r y da 2 2 <y, x 2, 3x 2 + 13y 2 > < 2x, 6y, 1> dx dy 2xy 6x 2 y + 3x 2 + 13y 2 dx dy [ x 2 y 2x 3 y + x 3 + 13y 2 x 2 dy 26y 2 2y + 8 dy (26/3)y 3 1y 2 + 8y 3 168. d. Let C be the boundary of, with orientation induced from the orientation n of given in (b). Check tokes theorem by computing C F dr and curl F d nd seeing that you get the same number for both. Hint: You can parametrize C by using the parametrization of in (a). If is the plane region corresponding to by this parametrization, then C is parametrized by the boundary of.
First compute the flux integral. curl F i j k / x / y / z y x 2 4z + y 2 x 2 ( (4z + y 2 x 2 )/ y (x 2 )/ z)i ( (4z + y 2 x 2 )/ x (y)/ z)j + ( (x 2 )/ x (y)/ y) k <2y, 2x, 2x 1>. Let be the rectangle x 2, y 3. From (2b) we get curlf d curlf (r(x, y)) r x r y da <2y, 2x, 2x 1> < 2x, 6y, 1> da 2 4xy 12xy + 2x 1 dx dy 8x 2 y + x 2 x x2 x dy 32y + 2 dy 16y 2 + 2y 3 138. Now, we compute the line integral. ince is parametrized by the rectangle using the function r(x, y), the boundary is parametrized by, using the restriction of r(x, y) to the edges of. This gives 4 separate pieces to s we go counter-clockwise around, starting at (, ): C 1, parametrized by r 1 (x) r(x, ) < x,, x 2 >, x goes from to 2 C 2, parametrized by r 2 (y) r(2, y) < 2, y, 4 + 3y 2 >, y goes from to 3 C 3, parametrized by r 3 (x) r(x, 3) <x, 3, x 2 + 12>, x goes from 2 to C 4, parametrized by r 4 (y) r(, y) <, y, 3y 2 >, y goes from 3 to We do the integrals one at a time and add the results: 2 F dr F (r 1 (x)) r 1(x) dx C 1 2 2 <, x 2, 3x 2 > <1,, 2x> dx 6x 3 dx (6/4)x 4 2 24.
F dr C 2 F (r 2 (y)) r 2(y) dy <y, 4, 12 + 13y 2 > <, 1, 6y> dy 4 + 72y + 78y 3 dy 4y + 36y 2 + (39/2)y 4 3 1915.5. C 3 F dr 2 2 2 F (r 3 (x)) r 3(x) dx <3, x 2, 3x 2 + 117> <1,, 2x> dx 3 + 6x 3 + 234x dx 3x + (6/4)x 4 + 117x 2 2 498. F dr C 4 3 3 3 F (r 4 (y)) r 4(y) dy <y,, 13y 2 > <, 1, 6y> dy 78y 3 dy (39/2)y 4 3 1579.5. F dr 24 + 1915.5 498 1579.5 138. 3. Let F be the vector field F (x) x x 3. a. Compute div F. Explicitly, x <x, y, z>, x (x 2 + y 2 + z 2 ) 1/2., and x F (x) < (x 2 + y 2 + z 2 ), y 3/2 (x 2 + y 2 + z 2 ), z 3/2 (x 2 + y 2 + z 2 ) divf [ x x (x 2 + y 2 + z 2 ) 3/2 + [ y + [ y (x 2 + y 2 + z 2 ) 3/2 z 3/2 >, z (x 2 + y 2 + z 2 ) 3/2.
We calculate the partial derivatives using the quotient rule: [ x x (x 2 + y 2 + z 2 ) 3/2 (x2 + y 2 + z 2 ) 3/2 1 (3/2)(x 2 + y 2 + z 2 ) 1/2 2x x (x 2 + y 2 + z 2 ) 3 (x2 + y 2 + z 2 ) 3/2 3(x 2 + y 2 + z 2 ) 1/2 x 2 (x 2 + y 2 + z 2 ) 3 (x2 + y 2 + z 2 ) 1/2 ((x 2 + y 2 + z 2 ) 3x 2 ) (x 2 + y 2 + z 2 ) 3 2x2 + y 2 + z 2 (x 2 + y 2 + z 2 ) 5/2. witching x with y and x with z gives [ y y (x 2 + y 2 + z 2 ) [ 3/2 z z (x 2 + y 2 + z 2 ) 3/2 x2 2y 2 + z 2 (x 2 + y 2 + z 2 ) 5/2 x2 + y 2 2z 2 (x 2 + y 2 + z 2 ) 5/2 and thus div F x2 2y 2 + z 2 (x 2 + y 2 + z 2 ) 5/2 + x2 2y 2 + z 2 (x 2 + y 2 + z 2 ) 5/2 + x2 + y 2 2z 2 (x 2 + y 2 + z 2 ) 5/2 x2 2y 2 + z 2 + x 2 2y 2 + z 2 + x 2 + y 2 2z 2 (x 2 + y 2 + z 2 ) 5/2. b. Compute F d, where a is as in (1) the sphere of radius a, center (,, ), and with orientation the outward normal vector. ince the radial vector x is perpendicular to the sphere, we have n x x
ince x a for x in, we have F d F nd a x x x 3 x d x x x d 4 x a 2 x d 4 x a 2 x d 4 1 a d 2 1 a 2 d 1 a 2 4πa2 (from (1b)) 4π. c. (You might want to wait until after class on Monday for this one, but try it before if you are a thrill-seeker) Let be the ellipsoid 5x 2 + 11y 2 + 17z 2 123, oriented with the outward normal. Use the divergence theorem and (b) to compute F d. Hint: Take a small enough so that is inside of nd let D be the solid region between nd. Use the divergence theorem: The boundary of D is D where means the sphere of radius a with the inward pointing normal. We have F d F d F d a F d D div F dv (the divergence theorem) D (since div F ). F d F d 4π. 4.(You might want to wait until after class on Monday for this one, but try it before if you are a thrill-seeker) Let 1 be the paraboloid z 1 x 2 y 2, z, oriented with the upward normal. Let 2 be the paraboloid z 2 2x 2 2y 2, z, also oriented with the upward normal. a. Let F be a vector field on 3. Use tokes theorem to show that curlf d curlf d. 1 2
Hint: Note that 2 and 1 both have the same boundary curve, x 2 + y 2 1, z. Let C be the common boundary curve of 1 and 2. By Two applications of tokes theorem curlf d F dr curlf d. 1 C 2 b. Let F be a vector field on 3. Use the divergence theorem to show that curlf d curlf d. 1 2 Hint: div(curl F ). Note that 2 lies above 1, meeting along their boundary curve, and consider the solid region D between 1 and 2. Let 1 be the surface 1 oriented with the downward pointing normal. ince 2 is the top part of D and 1 is the bottom, we have D 2 1, because the normal vector on D pointing away from D is the upward normal on 2, but the downward one on 1. Following the hint, we use the divergence theorem, using also that div(curlf ) : curlf d curlf d curlf d 2 1 D div (curlf ) dv and thus curlf d 1 curlf d. 2 D