# 7a3 2. (c) πa 3 (d) πa 3 (e) πa3

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1 1.(6pts) Find the integral x, y, z d S where H is the part of the upper hemisphere of H x 2 + y 2 + z 2 = a 2 above the plane z = a and the normal points up. ( 2 π ) Useful Facts: cos = 1 and ds = ±a sin φ a cos θ sin φ, a sin θ sin φ, a cos φ da. As 3 2 part of the problem, you need to decide which sign you need. (a) 7a3 2 (b) 7a3 2 (c) πa 3 (d) πa 3 (e) πa3 2 πa2 2 2.(6pts) Let z = f(x, y) and suppose f(2, 1) = 3. Suppose x = g(u, t) and y = h(u, t) with g( 1, 3) = 2 and h( 1, 3) = 1. hen z is a function of u and t. Find z at the point u ( 1, 3) if gradf = 4, 5 at (2, 1), gradf = 1, 1 at ( 1, 3) and g u (u, t) = 3 at ( 1, 3), h u (u, t) = 2 at ( 1, 3). Which number below is z u (u, t) at ( 1, 3)? (a) (b) 2 (c) 1 (d) 3 (e) 4

2 3.(6pts) Compute origin. D y da where D is the upper half of the disk of radius a centered at the (a) 2a3 3 (b) a3 3 (c) 2a2 3 (d) 4a2 3 (e) 4a3 3 4.(6pts) he two level surfaces f(x, y, z) = x 2 y xyz+z 2 = 7 and g(x, y, z) = x 2 +y 2 +z 2 = 14 intersect at the point (3, 1, 2). Which equation below is an equation for the tangent line to the curve of intersection at the point (3, 1, 2)? (a) t 1, 1, + 1, 1, 2 (b) (1 t) 3, 1, 2 + t 1, 1, 1 (c) 3, 1, 2 + t 1, 1, 1 (d) (1 t) 3, 1, 2 + t 2, 1, 2 (e) 3, 1, 2 + t 2, 1, 2

3 5.(6pts) Which function below is a potential function for the field F = yz 2 + 2xyz, xz 2 + x 2 z, 2xyz + x 2 y 2z (a) f(x, y, z) = xyz 2 + x 2 yz 1 2 z4 (b) f(x, y, z) = x 3 y 2 z + x 2 xyz 2 (c) f(x, y, z) = 2xyz xy2 z y 2 (d) f(x, y, z) = xyz x3 y 2 z + x 2 (e) f(x, y, z) = xyz 2 + x 2 yz z 2 6.(6pts) Find the area of the parallelogram with vertices (1, 2), (3, 5), (4, 3) and (6, 6). (a) 8 (b) 7 (c) 11 (d) 1 (e) 9

4 7.(6pts) Find the directional derivative of f(x, y) = x 2 xy + y 3 at the point (2, 1) in the same direction as the vector u = 1, 1. (a) 1 1 3, 1 (b) (c) 2 (d) 1 1 1, 3 (e) 2 8.(6pts) Maximize the function x 2y + z subject to the requirement that the points lie on the surface x 2 + 4y 2 + z 2 = 3. (a) 3 (b) he function has no maximum value on the surface. (c) 4 (d) 3 2 (e) 2

5 9.(6pts) Which iterated integral below gives the same number as the iterated integral y x5 + 1 dx dy (a) 2 x 4 /8 x5 + 1 dy dx (b) 2 8x 4 x5 + 1 dy dx (c) 16 8x 4 x5 + 1 dy dx (d) x 4 /8 2 x5 + 1 dy dx (e) 16 x 4 /8 x5 + 1 dy dx 1.(6pts) Find the integral F dr where C is the square in the xy-plane starting at the C origin, going to (, 2), then to (2, 2), then to (2, ) and finally back to the origin, and F = 8 x2 xy, y (a) 3 (b) 2 (c) 3 (d) 4 (e)

6 11.(6pts) Let F = x, 2y, x 2 + z. Compute F ds over the part of the surface x 2 + y 2 + S zx 2 + zy 2 + z 3 = 9 which lies above the xy plane. Note that S together with the disk D x 2 + y 2 9, z = bounds a solid E. ry integrating over a different surface. (a) 49π2 2 (b) 81π 4 (c) (d) π 3 (e) 7π 4 12.(6pts) Let S be the surface r(u, v) = uv 2, uv, u 2 v and note that r(2, 1) = 2, 2, 4. Which equation below is an equation of the tangent plane to S at 2, 2, 4? (a) 2x 6y + z = 4 (b) x + y + 4z = 2 (c) 2x + y + 2z = 14 (d) x y + 4z = 16 (e) 2x 2y + z =

7 13.(6pts) Find the area of the piece of the cylinder over y = x 3, x 1 above the plane z = and below the graph of the cylinder z = 36x 3. (a) π 3 36 (b) 18π (c) 2 3( 1 1 ) (d) 12π ( 1 ) (e) (6pts) If C is the curve r(t) = (sin 2t) ( 4 t 2, t π ), 2 + sin(t), t π 2 2 find f dr where f(x, y, z) = x 2 y 3 + zy + xy z. C (a) 1 (b) (c) π (d) 16 5 (e) 2

8 15.(6pts) Let r(u, v) = u 2 + uv, v 2 + uv. Compute the Jacobian of this transformation. (a) u 2 + v 2 4uv (b) 4uv + 2u 2 + 2v 2 (c) 4uv 2u 2 + 2v 2 (d) 4uv + 2u 2 2v 2 (e) (u 2 + v 2 )uv 16.(6pts) Find the moment about the yz plane of a thin sheet of unit density bent in the shape of a surface r(u, v) = u 3, v 4, uv for 1 u 1 and 1 v 1. Suppose is the square with vertices (, ), (, 1), (1, ) and (1, 1). (a) u 3 u 6 + v 8 + u 2 v 2 da (b) u 3 16v 8 + 9u u 4 v 6 da (c) (e) v 4 16v 8 9u u 4 v 6 da v 4 16v 8 + 9u u 4 v 6 da (d) u 3 16v 8 9u u 4 v 6 da

9 17.(6pts) Find the area of the region enclosed by the curve x = sin(2t), y = sin(t), t π. Fact: sin(2t) = 2 sin(t) cos(t), cos(2t) = cos 2 (t) sin 2 (t). Hint: Use Green s heorem. (a) 8 2π 5 (b) 2π 3 (c) 2π 1 5 (d) 4 3 (e) 4π (6pts) Which number below is S xyz, xyz, xyz ds where S consists of the six faces of the cube with sides of length 2 in the first octant and with one vertex at the origin and with normal vector pointing out of the cube? (a) 4 (b) 6 (c) 12 (d) 24 (e) 1

10 19.(6pts) Let C be the intersection of the cylinder over the triangle with vertices (, ), (1, ) and (, 1) with the sphere x 2 +y 2 +z 2 = 9. Orient counterclockwise. Which integral below is equal to y, z, x dr? (a) (d) C x y 9 x2 y 1 da(b) 2 x y 9 x2 y + 2 da(e) 2 x + y 9 x2 y 1 da(c) 2 x y 9 x2 y da x y 9 x2 y 2 1 da 2.(6pts) Let r(t) = t 2, (t 1) 3, t 2. Which parametrized surface below is the result of rotating this curve about the y axis? (a) t 2, (t 1) 3 cos(θ), t 2 sin(θ) ; t 2, θ 2π (b) t 2 cos(θ), (t 1) 3, t 2 sin(θ) ; t 2, θ 2π (c) t 2, (t 1) 3 cos(θ), t 2 sin(θ) ; t 2, θ π (d) t 2 cos(θ), (t 1) 3, t 2 sin(θ) ; t 2, θ π (e) t 2 cos(θ), (t 1) 3 cos(θ), t 2 sin(θ) ; t 2, θ π

11 21.(6pts) Which point below is a local maximum for the function f(x, y) = 2x 3 + 6xy 2 3y 3 15x. Note f xx = 12x, f xy = f yx = 12y and f yy = 12x 18y. he function f has exactly four critical points listed as answers (a)-(e). (a) (5, ) (b) (3, 4) (c) ( 3, 4) (d) he function had no local maxima. (e) ( 5, ) 22.(6pts) Find the length of the curve given by r(t) =,, and 72, 3, t3, 1 2 t, 1 2 t2 between the points (a) 34 (b) 5 (c) 19 (d) 75 (e) 86

12 23.(6pts) Which number below is the cosine of the angle of intersection of the two planes 2x + 3y + z = 1 and 3x + 2y z = 4. (a) 4 5 (b) 3 16 (c) (d) 2 3 (e) he two planes do not intersect. 24.(6pts) Let F = x + e sin(yz), y + sin(x 2 + z), cos(xyz). hen div F is a function, grad(div F) is a field and so curl ( grad(div F) ) is a field. Which field below is it? (a) x, 2y, z (b) 1, 1, sin(xyz) (c) 1, 2, 1 (d) x, y, z (e),,

13 25.(6pts) Compute the divergence of F, F where F = f with f(x, y, z) = x 2 z +xyz +y 3. (a) 2xz + xz (b) 3xz + yz + 3y 2 (c) (d) 3xz + yz + 3y 2 + x 2 + xy (e) 6y + 2z

14 1. Solution. In spherical coordinates, H is given by ρ = a, θ 2π, φ π 3, so H is parametrized by r(θ, φ) = a cos θ sin φ, a sin θ sin φ, a cos φ so a normal vector is given by i j k r θ r φ = det a sin θ sin φ a cos θ sin φ = a cos θ cos φ a sin θ cos φ a sin φ a 2 cos θ sin 2 φ, a 2 sin θ sin 2 φ, a 2 sin φ cos φ = a sin φ a cos θ sin φ, a sin θ sin φ, a cos φ. he region in the θ-φ plane is given by θ 2π, φ π 3 Hence x, y, z d S = H a cos θ sin φ, a sin θ sin φ, a cos φ a sin φ a cos θ sin φ, a sin θ sin φ, a cos φ da = 2π π/3 2π a(sin φ)a 2 da = a 3 sin φ da = a 3 sin φ dφ dθ = a 3 cos φ a 3 2π 1 2. π/3 dθ = 2. Solution. Let r = g, h. hen g u (u, t) = f r u at ( 1, 3) so the answer is 4, 5 3, 2 = Solution. In polar coordinates, D is r a, θ π so y da = ( D a ) ( π ) r 2 dr sin(θ) dθ = a3 3 ( cos(θ) π ) = a3 2a3 ( 1 (1)) = 3 3 a π r sin(θ)r dθ dr =

15 4. Solution. f = 2xy yz, x 2 xz, xy + 2z and g = 2x, 2y, 2z so at the point in question f = 4, 3, 1 and g = 6, 2, 4 so a vector on the tangent line is f g = i j k det = 12 2, (16 6), 8 18 = 1, 1, 1 A parallel vector is 1, 1, Hence 3, 1, 2 + t 1, 1, Solution. f x = yz 2 + 2xyz so f = xyz 2 + x 2 yz + g(y, z). f y = xz 2 + x 2 z + g y = xz 2 + x 2 z so g y = and g(y, z) = h(z). f z = 2xyz + x 2 y + h = 2xyz + x 2 y 2z so h = 2z and h = z 2. f(x, y, z) = xyz 2 + x 2 yz z 2 i j k 6. Solution. (1, 2) + 2, 3 = (3, 5), (1, 2) + 3, 1 = (4, 3). Area is ± det 2 3 = 3 1,, 2 9 = 7 so 7.

16 7. Solution. First f = 2x y, 3y 2 x and f(2, 1) = 3, 1. he unit vector is the 1 1 direction of u is 1, 1 so D u f(2, 1) = 3, 1 1, 1 = 2 = Solution. If f = x 2y + z and g = x 2 + 4y 2 + z 2 then Lagrange says the maxima and minima occur at points where f = λ g and g = 3. Hence 1, 2, 1 = λ 2x, 8y, 2z or 1 x, y, z = 2λ, 1 ( ) 2 ( ) 2 ( ) 2 4λ, Hence g(x, y, z) = +4 + = 3 or 3 1 2λ 2λ 4λ 2λ 4λ = 3 2 or λ = ± 1 ( 2. Hence there are eight points ±1, ± 1 ). 2, ±1 he maximum occurs when all three terms are positive so f = 1 2( 1/2) + 1 = = Solution. Write the iterated integral as a double integral over the region D given by 4 8y x 2; y 2. Setting up the other way, 2 x 4 /8 x5 + 1 dy dx.

17 ( N 1. Solution. By Green s heorem Mdx + Ndy = S S x M ) da where S is the y square and F = M, N. he curve S is C so if we compute the double integral we will get the negative of the answer. We get () ( y) da = y dx dy = 2 y dy = y 2 2 = 4. S 11. Solution. S x dx 2y dy + (x 2 + z) dz + D x dx 2y dy + (x 2 + z) dz = where the normal vector to D points down. F = x, 2y, x 2 + z. o parametrize D use r(x, y) = x, y, with normal vector,, 1. x dx 2y dy + (x 2 + z) dz = D 3 2π ( 3 ) ( 2π ) x 2 da = r 2 cos 2 (θ)r dθ dr = r 3 dr cos 2 (t) dt = 81π 4. D E dv 12. Solution. r u = v 2, v, 2uv and r v = 2uv, u, u 2 so at u = 2, v = 1, i j k i j k r u r v = det v 2 v 2uv = det = (4 8), (4 16), (2 4) = 4, 12, 2 2uv u u So 4, 4, 2 x 1, y 2, z 4 = or 4x + 4y 2z = 4.

18 13. Solution. he surface is parametrized by r(x, z) = x, x 3, z over the region given by i j k x 1 and z 36x 3. he normal vector is det 1 3x 2 = 3x 2, 1, x 3 1 so 1dS = 1 + 9x4 da = 1 + 9x4 dz dx = 36 (x 3 ) 1 + 9x 4 dx. Let S 1 u = 1 + 9x 4 so du = 36x 3 dx so 36 x ( ) 1 + 9x 4 3 dx = u du = 1 1 = ( ) Solution. By the Fundamental heorem, the answer is f ( r(b) ) f ( r(a) ) ( π ) and r ( 2 π )) (, π ) 2 2, 2,, 3, r() =, π 2, 2. Now f so the answer is ( π) = π. ( r = f (,, 3) = ; f (r()) = f = = π 15. Solution. he Jacobian is det 2u + v u = (2u + v)(2v + u) (u)(v) = (4uv + 2u2 + 2v 2 + uv) (uv) = 4uv + 2u 2 + 2v 2.

19 i j k 16. Solution. he short answer is x ds. A normal vector is det 3u 2 v = 4v 4, 3u 3, 12u 2 v 3 S 4v 3 u and its length is 16v 8 + 9u u 4 v 6 so x ds = u 3 16v 8 + 9u u 4 v 6 da. S 17. Solution. r = 2 cos(2t), cos(t) he area is da =, x 2 cos(2t), cos(t) dt = A π π sin(2t) cos(t) dt = 2 sin(t) cos 2 (t) dt = 2 cos3 (y) π = ( ( 2/3) (2/3) ) = π 18. Solution. By the Divergence heorem, (yz + xz + xy) dx dy dz = (2z + 1) dz = 4 ( z 2 + z ) 2 = 24 S 2 2 xyz, xyz, xyz ds = (2yz + 2z + 2y) dy dz = U 2 F dv = (4z + 4z + 4) dz =

20 19. Solution. By Stoke s heorem, the answer is S curl F ds where S is the part of the sphere lying over the triangle. he part of the sphere we have is the graph of z = x 9 x2 y 2 over the triangle. So, d S = 9 x2 y, y 2 9 x2 y, 1 da 2 i j k det = 1, (1), 1 = 1, 1, 1 x y z y z x so we need to integrate x y 9 x2 y 1 da 2 over the triangle. 2. Solution. R(t, θ) = t 2 cos(θ), (t 1) 3, t 2 sin(θ) ; t 2, θ 2π

21 21. Solution. he critical points are located at the zeros of the gradient: f x = 6x 2 + 6y 2 15, f y = 12xy 9y 2. Since 12xy 9y 2 = 3y(4x 3y either y = or 4x 3y = so y = 4 3 x. Note f x = is the same as x 2 + y 2 = 25. If y =, x = ±5. If y = 4 3 x x x2 = 25 or 25 9 x2 = 25 so x = ±3 and the critical points are (5, ), ( 5, [ ) (3, 4) ] and [ ( 3, 4). ] [ ] fxx f he Hessian is yx 12x 12y 2x 2y = = 6 whose determinant f xy f yy 12y 12x 18y 2y 2x 3y is 4x 2 6xy 4y 2 = 2(2x 2 3xy 2y 2 ) so the type of critical point is determined by h(x, y) = 2x 2 3xy 2y 2. h(5, ) = 25 = h( 5, ) >, h(3, 4) = h( 3, 4) = = 24 <. Negative Hessian means saddle point so (pm5, ) are the only possibilities. f xx (5, ) = 6 > so (5, ) is local minimum; f xx ( 5, ) = 6 < so ( 5, ) is local maximum. 22. Solution. he parametrized curve is r, t 6. Hence the length is r (t) dt. Now r (t) = t 2, 12, t so r (t) = t t2 = t so r (t) dt = 1 3 t t 6 = Solution. he angle is the angle between the two normal vectors 2, 3, 1 and 3, 2, 1 and so cos θ =

22 24. Solution. he curl of a gradient is,, and we have a gradient. 25. Solution. F = f = 2xz + yz, xz + 3y 2, x 2 + xy so F = (2z) + (6y) + () = 6y + 2z.

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