# Math 233. Practice Problems Chapter 15. i j k

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1 Math 233. Practice Problems hapter ompute the curl and divergence of the vector field F given by F (4 cos(x 2 ) 2y)i + (4 sin(y 2 ) + 6x)j + (6x 2 y 6x + 4e 3z )k olution: The curl of F is computed as i j k curlf x y z 4 cos(x 2 ) 2y 4 sin(y 2 ) + 6x 6x 2 y 6x + 4e 3z 6x 2 i + (6 12xy)j + (6 + 2)k 6x 2 i + (6 12xy)j + 8k The divergence of F is divf 8x sin(x 2 ) + 8y cos(y 2 ) 12e 3z 2. Determine whether the vector field F(x, y, z) [5x 4 z]i + [4y 3 e 8z + 5 cos(5y)]j + [8y 4 e 8z + x 5 + 2]k is conservative, and if it is conservative find a potential function for it. olution: Yes, F is a conservative vector field, because denoting F Mi + Nj + P k we have M y N x, M z 5x4 P x and P y 32y3 e 8z N z In other words, curl(f). A potential function f(x, y, z) satisfies f(x, y, z) 5x 4 z dx x 5 z + 1 (y, z) and and f(x, y, z) f(x, y, z) [4y 3 e 8z + 5 cos(5y)] dy y 4 e 8z + sin(5y) + 2 (x, z) [8y 4 e 8z + x 5 + 2] dz y 4 e 8z + x 5 z + 2z + 3 (x, y) Putting these altogether a potential function is f(x, y, z) y 4 e 8z + x 5 z + sin(5y) + 2z + 3. Evaluate the line integral y 4 x ds where is the portion circle x 2 + y 2 16 starting at ( 4, ) and ending at (, 4).

2 olution: Parametrize by x 4 cos t, y 4 sin t for π t 3 2 π. Then ds 16 sin 2 t + 16 cos 2 t dt 4 dt and 3π/2 y 4 x ds 4 4 (sin 4 t)(4 cos t)(4) dt 496 3π/2 sin 5 t π 5 π (( 1)5 ) Evaluate the line integral (4x 7y 2 ) dx + 26xy dy where is the parabola y 2 x starting at (4, 2) and ending at (1, 1). olution: Parametrize the curve as x t 2 and y t where 2 t 1. Then dx 2t dt and dy dt. Thus the integral becomes (4xy 7y 3 ) dx + 26xy 2 dy [(4t 3 7t 3 )(2t) dt + 26(t 4 )(t)] dt 1 2t 4 dt 4t 5 4( 1) 5 4( 2) (a) Evaluate the line integral ending at (, 3, ). (b) Evaluate the line integral 2yx dx + xz dy + z dz where is the line segment starting at (1, 3, 2) and 2yx dx + xz dy + z dz where is the line segment starting at (, 3, ) and ending at (1, 3, 2) (note this is the same line integral as in (a), but with the starting and ending points of (c) reversed). olution: (a) A parametrization of a line segment starting at (x, y, z ) and ending at (x 1, y 1, z 1 ) is x x + at, y y + bt and z z + ct for t 1, where a x 1 x, b y 1 y and c z 1 z. Thus the line segment is given by x 1 1t, y 3, z 2 + 2t, t 1. Therefore, dx 1 dt, dy dt and dz 2 dt, and thus the integral becomes 2yx dx + xz dy + z dz 1 1 2(3)(1 1t)( 1 dt) + dt + ( 2 + 2t)(2 dt) [1t 1] dt (b) The answer is 5, because it is the negative of the answer in (a) since the orientation on the curve was reversed, but everything else remained the same. 6. (a) Find the work done by the force field F(x, y) 6yi + 5xy 2 j along the parabola x y 2 starting at (25, ) and ending at (25, 5).

3 (b) Find the work done by the force field F(x, y) as in (a), but along the straight line from (25, ) to (25, 5). olution: (a) Parametrize the parabola by x t 2 and y t for t 5. Then dr 2t, 1 dt and the work is given by W F dr 6y, 5xy 2 2t, 1 dt 5 5 6t, 5t 4 2t, 1 dt (12t 2 + 5t 4 ) dt 4t 3 + 1t 5 5 [2(4)(5) 3 + 2(1)(5) 5 ] 725 (b) Parametrize the line segment by x 25 and y t for t 5. Then dr, 1 dt and the work is given by W F dr 6y, 5xy 2, 1 dt 5 5 6t, 5(25)t 2, 1 dt (125t 2 ) dt t3 2(125)(5) 3 /3 3125/3 7. (a) Find a potential function for the following conservative vector field F(x, y) ( 16x 7 y 7 6 ) i + 14x 8 y 6 j (b) Use the fundamental theorem of line integrals to evaluate F dr where F is the vector field in (a) and is the portion of the parabola y x 2 starting at (, ) and ending at (1, 1). olution: (a) First, f(x, y) (16x 7 y 7 6 ) dx 16 8 x8 y 7 6x + (y) 2x 8 y 7 6x + (y) similarly, f(x, y) 14x 8 y 6 dy 14 7 x8 y 7 + (x) 2x 8 y 7 + (x) Putting these together implies f(x, y) 2x 8 y 7 6x +. (b) Using the potential function from (a) we have F dr 2x 8 y 7 6x (1,1) (,)

4 8. Verify that the vector field F(x, y, z) (4y + 3z)i + (4x 4y 5z)j + (3x 5y)k is conservative. (b) Evaluate the integral F dr along any piecewise smooth curve starting at (2, 1, ) and ending at (1, 1, 2) where F is the vector field in (a). olution: (a) Because M, N and P have continuous partial derivatives, it suffices to show that curlf, which we now do. i j k curlf x y z ( ())i + (3 3)j + (4 4)k. 4y + 3z 4x 4y 5z 3x 5y (b) A potential function f for F satisfies f(x, y, z) (4y + 3z) dx 4xy + 3xz + 1 (y, z) (4x 4y 5z) dy 4xy 2y 2 5yz + 2 (x, z) (3x 5y) dz 3xz 5yz + 3 (x, y) putting this together we find f(x, y, z) 4xy + 3xz 2y 2 5yz is a potential function for F. Thus using the the Fundamental theorem of line integrals we obtain F dr 4xy + 3xz 2y 2 5yz (1,1, 2) (2, 1,) [4(1)(1) + 3(1)( 2) 2(1) 2 5(1)( 2)] 6 ( 1) 16 [4(2)( 1) + 3(2)() 2( 1) 2 5( 1)()] 9. Evaluate the line integral (5y + x 2 ) dx + (5x e y2 7y) dy where is the bottom half of the circle x 2 + y 2 9 starting at ( 3, ) and ending at (3, ). Hint: if the integral is path independent, find a simpler path. olution: Notice that this is a path independent line integral since y (5y + x2 ) 5 ( ) 5x e y2 7y. x However, it is hard to find a potential function because of the difficulty in finding an antiderivative for e y2. Therefore, we choose a different path, namely, x t, y, for t 3 to t 3. For this, dx dt and

5 dy dt. Then the line integral becomes (5y + x 2 ) dx + (5x e y2 7y) dy t3 3 (5() + t 2 ) dt + dt 3 2(33 ) (a) Is the line integral 2yx dx + xz dy + z dz path independent? (b) Evaluate 2yx dx+xz dy+z dz where is the line segment starting at ( 2, 2, 3) and ending at ( 6, 2, 3). olution: (a) No, the line integral is not path independent, because the curl of F 2yxi + xzj + zk is not the zero vector, since, e.g. x (xz) (2yx) z 2x y (b) ince the line integral is not path independent, we will evaluate it directly. First, a parametrization of a line segment starting at (x, y, z ) and ending at (x 1, y 1, z 1 ) is x x + at, y y + bt and z z + ct for t 1, where a x 1 x, b y 1 y and c z 1 z. Thus the line segment is given by x 2 4t, y 2, z 3 + 6t, t 1. Therefore, dx 4 dt, dy dt and dz 6 dt, and thus the integral becomes 2yx dx + xz dy + z dz 1 1 2(2)( 2 4t)( 4 dt) + dt + ( 3 + 6t)(6 dt) [1t + 14] dt Use Green s theorem to compute the work done by the force field F(x, y) 3y 2 i + 11xyj where is the boundary of the region bounded by the graphs of y, y x and x 2 oriented in a counter clockwise direction.

6 olution: We will use Green s theorem to compute the work integral ( F dr x (11xy) ) y (3y2 ) da (11y 6y) da 2 x y2 2 (5y) dy dx x 5 2 x dx x2 5 dx 12. Use Green s theorem to evaluate the integral 5xy dx + (4x 2 3 sin y) dy where is the boundary of the trapezoid with vertices at (, ), (2, ), (2, 1) and (, 11), and is oriented in a counterclockwise direction. olution: A sketch of is given for reference. y x With reference to the graph above, the upper boundary of the trapezoidal region is the line y 11 5x since it has y-intercept (, 11) and slope. Therefore, the region is given by x 2 and y

7 11 5x. Using Green s theorem, we find 5xy dx + (4x 2 3 sin y) dy ( x (4x2 3 sin y) ) y (5xy) da 2 11x 2 11x 2 33x2 2 (8x 5x) dy dx ( 2 3x dy dx 3xy (33x 15x 2 ) dx 15x3 3 2 y11x y ) dx 13. onvert the following vector-valued function to rectangular form. r(u, v) v cos ui + v sin uj + 6v 2 k olution: In this case x v cos u and y v sin u and so x 2 + y 2 v 2 and so z 6v 2 means which is the equation of a paraboloid. z 6x 2 + 6y A surface is given by the vector-valued function r(u, v) ui + vj + uvk, u, v Find the equation of the tangent plane to the surface at the point (6, 54, 18) olution: When u 6 and v 54 we have r u 1,, /6 1,, 3/2 and rv, 1, 1 2 The cross-product will provide a normal vector for the tangent plane: i j k r u r v /2, 1/6, /54, 1, 1/6 We may choose any convenient non-zero multiple of the cross-product for our normal vector, so we choose n 9, 1, 6. The desired plane thus goes through (6, 54, 18) and has normal vector 9, 1, 6 so it has equation 9(x 6) + 1(y 54) 6(z 18) which can be written as 9x + y 6z.

8 15. Use an integral to find the area of the surface given by r(u, v) 5 cos ui + 5 sin uj + vk, u 2π, v 7 olution: To find the surface area, we compute i j k r u r v sin u 5 cos u 5 cos u, 5 sin u,. 1 Then r u r v 5 2 (cos 2 u + sin 2 u) 5 and so the surface area is given by 2π 7 5 dv du 2π(7)(5) 7π To check this answer, we observe that this represents the portion of the cylinder x 2 + y 2 25 that is centered on the z-axis and between the planes z and z 7. The lateral surface area of a cylinder with radius r and height h is given by 2πrh and in this case r 5 and h 7 so the lateral surface area is 2π(5)(7) 7π as above. 16. Evaluate the surface integral (7x + 3y + z) d where is the surface z 5 x + 4y over the rectangle 3 x 3, y 5. olution: First d 1 + ( 1) 2 + (4) 2 da 18 da. The surface integral is then (7x + 3y + z) d (7x + 3y + 5 x + 4y) 18 dy dx (5 + 6x + 7y) 18 dy dx (5 + 6x + 7y) 18 dy dx dy dx (6)(1)(5) (6x + 7y) 18 dy dx 17. Find the mass of the cylindrical surface x 2 + y 2 49, for z 5 whose density at each point is f(x, y, z) 3x 2 z + 3y 2 z. olution: Write the surface parametrically r(u, v) 7 cos ui + 7 sin uj + vj u 2π, v 5 Then r u 7 sin u, 7 cos u,, and r v,, 1. Then r u r v 7 cos u, 7 sin u,

9 and then r u r v 7 and so d 7 dv du. The density is f(x, y, z) 3(x 2 + y 2 )z 147v. The mass is then given by f(x, y, z)d 2π 5 129v dv du 5 129πv π 18. For this question we consider the vector field F 2xi + 2yj + 4zk (a) Find the flux of F through the surface z and x 2 + y 2 1 oriented with downward unit normal. (b) Find the flux of F through the surface z 1 x 2 y 2 where z and the surface is oriented with upward unit normal. olution: (a) For this surface Nd,, 1 da and so the flux is 2x, 2y, (4)(),, 1 da da (b) For this surface z 1 x 2 y 2 where x 2 + y 2 1 then N d 2x, 2y, 1 da and so the flux is given by F N d 2x, 2y, 4z 2x, 2y, 1 da (4x 2 + 4y 2 + 4(1 x 2 y 2 )) da since z 1 x 2 y 2 2π 2π 1 1 (4r 2 + 4(1 r 2 ))r dr dθ [ ] 1 (r 3 + 4r) dr π r 4 + 4r 2 π + 4π 4π 19. Use the divergence theorem to evaluate the integral (F N)d where is the closed surface bounded by the planes y 2 and z 4 x and the coordinate planes, and, as usual, is oriented with outward unit normal, and where F(x, y, z) 7xyi + 4xzj + (y 3yz)j

10 olution: To apply the divergence theorem, we first find div F 7y + 3y 4y. Now let Q be the solid enclosed by the surface. According to the divergence theorem, the outward flux through the surface is (F N)d divf dv Q x y dz dy dx 4y(4 x) dy dx 2y 2 (4 x) (32 8x) dx 32x 8 x2 2 4 y2 y 64 dx 2. Use the divergence theorem to find the outward flux (F N)d where F(x, y, z) x 2 i + 4yj 5zk, and is surface of the solid bounded by the graphs of x 2 + y 2 16, z and z 9 oriented with outward unit normal. olution: To apply the divergence theorem, we first find div F 2x x Now let Q be the solid enclosed by the surface. According to the divergence theorem, the outward flux through the surface is (F N)d divf dv ( 1 + 2x) dv Q 2π 4 9 2π 4 9 Q ( 1 + 2r cos θ)r dz dr dθ ( 1)r dz dr dθ + 2π 4 9 ( 1)(Area of Q) + ( 1)(9)(4 2 )π 144π (2r cos θ)r dz dr dθ 21. Use the divergence theorem to find the outward flux (F N)d

11 where F(x, y, z) 4xi+(e z 3y)j+(5z+sin(6x))k, and is surface of the sphere whose equation is x 2 +y 2 +z 2 81 oriented with outward unit normal. olution: To apply the divergence theorem, we first find div F According to the divergence theorem, the outward flux through the surface is ( ) 4 (F N)d divf dv (6)(Volume of Q) (6) 3 π π Q where the volume of Q was given by 4 3 πr3 where r 9 is the radius of the sphere. 22. Let be the closed surface the encloses the solid bounded by the graphs z z 2 + y 2 and z 9 oriented with outward unit normal. Use the divergence theorem to find the flux of the vector field F(x, y, z) (6x xz)i + 13e z j (7z 2 14y)k over the surface. olution: First, the divergence of F is and so the divergence theorem says (F N) d Note. It was clear above that 2π π div F 12x + 23z 14z 12x + 9z Q 2π 3 9 2π 3 9 2π (12x + 9z) dv r 2 (12r 2 cos θ + 9zr) dz dr dθ r 2 9zr dz dr dθ + 2π 3 9 9zr dz dr dθ + r 2 9rz 2 z9 3 2 dr π (729r 9r 5 ) dr zr 2 ] 9r π [ 729r 2 2π 6 13 one can note that any integral of the form 3 9r 2 cos θ dz dr dθ since Q (ax + by) dv when Q is a closed bounded solid that is symmetric about the z-axis. r 2 (12r 2 cos θ) dz dr dθ 2π cos θ dθ. More generally,

12 23. Use tokes theorem to evaluate the line integral F dr where F is the vector field F (4e x2 4y)i + (16 sin(y 2 ) + 3x)j + (4y 4x e z )k and is the triangle with vertices (1,, ), (, 2, ), (,, 5) oriented counterclockwise when viewed from above. olution: First, the curl of F is computed as i j k curlf x y z 4e x2 4y 16 sin(y 2 ) + 3x 4y 4x e z 4i 4j + 7k. Now, lies on a plane z ax + by + d; because contains (,, 5) we know d 5, because contains (1,, ) we know a(1) + 5 and so a 1/2 because contains (, 2, ) we know b(2) + 5. This implies b /2, and so one possible equation of the plane is z 1 2 x 5 2 y + 5 Now write the plane as 1 2 x y + z 5. Writing G(x, y, z) 1 2 x z 5 we find 1 N d G da 2, 5 2, 1 da Observe that the plane lies above the region in the xy-plane which is a triangle with vertices (,, ), (1,, ), (, 2, ). Now applying tokes theorem we find 1 F dr 4, 4, 7 2, 5 2, 1 da [(4)(1/2) + (4)(5/2) + (7)(1)] da 19 da (19)(Area of ) (19)(1)(2) Use tokes theorem to evaluate the line integral F dr where F is the vector field F(x, y, z) xyzi + (6y + 7x)j + (z e 5z )k where is the first-octant portion of z x 2 over the circle x 2 + y 2 a 2, where a > is a constant, and is the boundary of oriented counterclockwise when viewed from above. olution: First, the curl of F is curlf i j k x y z xyz 6y + 7x z e 5z i + xyj + (7 xz)k.

13 Writing the surface as G(x, y, z) z x 2 then N d 2x,, 1 da and by tokes theorem, the line integral is F dr, xy, 7 xz 2x,, 1 da (7 x 3 ) da since z x 2 π/2 a π/2 a 7πa2 4 2a5 15 (7 r 3 cos 3 θ) r dr dθ (7r r 4 cos 3 θ) dr dθ 25. (a) Use tokes theorem to evaluate the line integral F dr where F is the vector field F (4x 2 e z + 5z 4 y + 7y)i + (2x + 3y sin(z 3 ))j + (xy ye z )k and is the boundary of where is the capped cylindrical surface which is the union of two surfaces, a cylinder given by x 2 + y 2 16, z 5, and a hemispherical cap given by x 2 + y 2 + (z 5) 2 16, z 5, and is oriented in a counterclockwise orientation when viewed from above. For this, you can replace the surface with a more convenient oriented surface with the same boundary because for both surfaces 1 and 2 we have curlf N d F dr curlf N d 1 2 (b) heck your answer by evaluating the line integral directly. olution: (a) We will evaluate curlf N d where is the surface z that is enclosed by the circle x 2 + y 2 16 and oriented with upward unit normal k. (This is just Green s theorem). Thus curlf N d [2 (5z 4 + 7)] da () da [since z ] ()(Area of ) ()(16π) 8π (b) The curve can be parametrized by r(t) 4 cos ti + 4 sin tj + k for t 2π. Then r (t) ( 4 sin ti + 4 cos tj + k). On this curve noting z, F becomes F(r(t)) 4(16 cos 2 t) + 7(4 sin t), 2(4 cos t), 16 cos t sin t e sin t and so F(r(t)) r (t) 256 cos 2 t sin t 112 sin 2 t + 32 cos 2 t

14 The line integral is F dr 2π which agrees with the answer in (a) as it should. ( 256 cos 2 t sin t 112 sin 2 t + 32 cos 2 t) dt ( )π 8π

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