x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the


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1 1.(8pts) Find F ds where F = x + ye z + ze y, y + xe z + ze x, z and where T is the T surface in the pictures. (The two pictures are two views of the same surface.) The boundary of T is the unit circle in the xy plane. Let D be the unit disk in the xy plane. Then T D is the boundary of a solid E whose volume is π. The orientation on T is outward with respect to E. You are looking at the picture from underneath. This problem will require some thought since all you know about T and E are what you see in the picture plus the volume of E. (a) 6π (b) π (c) 6π (d) π(e e) (e) π
2 .(8pts) Let T be the surface r(u, θ) = u, u cos(θ), u sin(θ) for u 1, θ π. This is the surface you get by rotating y = x, x 1, around the y axis. If the density at a point on this surface is given by 4x 5 + 4x 3, find the mass. Remark: The density has been chosen so that a simple substitution can be used to evaluate the integral you should get. (a) 4π (b) π(4) 3 (c) π (d) π 8 (e) π ( 4 )
3 3.(8pts) Let C be the curve parametrized by r(t) = cos t, sin t, 1. Stokes theorem tells us that for a differentiable vector field F F dr = (curl F) ds C for any oriented smooth surface S. The surfaces below are possible choices for S where the vectors indicate the direction of the normal field. Which of the following is an appropriate choice for S? The curve C is drawn in bold in each picture and the positive directions on the axes are labeled. Hint: Be mindful of the orientation of C! S (a) (b) (c) (d) (e)
4 4.(7pts) We are asked to find absolute maximum and minimum values of f(x, y, z) with respect to the constraints g(x, y, z) = 1 and h(x, y, z) = e. We are given that the five points (1, 1, 1), (, 1, 1), (, 1, 1), (1, 1, 1) and ( 1, 1, 1) satisfy both constraint equations. Each of the five points satisfies an additional condition: f(1, 1, 1) = g(1, 1, 1) + 3 h(1, 1, 1) f(, 1, 1) = 6 g(, 1, 1) f(, 1, 1) ( ( g(, 1, 1) h(, 1, 1) ) = f( 1, 1, 1) ( ( g( 1, 1, 1) h( 1, 1, 1) ) f(1, 1, 1) = Which point can not be an absolute extremum? (a) (1, 1, 1). (b) (1, 1, 1). (c) ( 1, 1, 1). (d) (, 1, 1). (e) (, 1, 1). 5.(7pts) Compute the angle between the vectors 1, 1, and, 1, 3. (a) π/4 (b) π (c) π/ (d) arccos ( ) 1 (e) arccos 8 ( ) 1 8
5 6.(7pts) Compute the flux F ds, where F = 4y, xy, xz and S is the surface z = ye x, x 1, y 1, with the upward orientation. S (a) e (b) 1 e (c) e 1 (d) e (e) 1 + e 7.(7pts) Find F dr where F = x, z, y and C is parametrized by C r(t) = cos t, t, sin t, t π. (a) π + (b) π 1 (c) π (d) 1 (e) π
6 8.(7pts) Compute x(t) dt where x(t) = sin(t)i + e t k. (a) cos t + e t + C (c) ( cos t + C 1 )i + (e t + C )k (b) ( cos t)i + e t k (d) ( cos t + C)i + Cj + (e t + C)k (e) ( cos t + C 1 )i + C j + (e t + C 3 )k 9.(7pts) If f(x, y) = sin(xy) find the directional derivative D u f at the point (x, y), where 1 1 u =,. (a) (d) x + y cos(xy) x + y sin(xy) (b) x + y sin(xy) (c) x y cos(xy) (e) x + y cos(xy)
7 1.(7pts) Determine the surface area of the part of z = xy that lies inside the cylinder x +y = 1. (a) π 3 ( 3 + 1) (b) π 3 ( 3 1) (c) π (d) π 3 ( 3 + 1) (e) π 3 ( 3 1) 11.(7pts) Find F ds where F = x, yz, z and D is the surface of the cube E with D vertices: (,, ), (,, ), (, 3, ), (,, 4), (, 3, ), (, 3, 4), (,, 4), and (, 3, 4), so x, y 3 and z 4. (a) (b) 4 (c) 48 (d) 36 (e) 4
8 1.(7pts) Which of the following is perpendicular to both a and b if a b = 1 and a b,,. (a) (a b) b (b) (a b) a (c) (a + b) (a b) (d) (a b)(a b) (e) a + b 13.(7pts) Compute the curl of the vector field F(x, y, z) = e x sin(y), e y cos(z), e x. (a) e y sin(z), e x, e x cos(y) (c) e y sin(z), e x, e y cos(y) (b) e x sin(z), e x, e x cos(y) (d) e y sin(z), e x, e x cos(y) (e) e y sin(z), e x, e x cos(y)
9 14.(7pts) Find the center of mass of the thin plate D which has the shape of a halfdisk consisting of the region below the semicircle bounded by x + y = 9 and above the xaxis. Assume that D has density ρ(x, y) = x + y and mass 9π. ( ) ( (a), (b), π ) ( ) ( ) ( ) 9 3 (c), (d), (e), 9π 3 π 3π π 15.(7pts) Let f(x, y) = x + 6xy 3y. Find and classify all critical points. (a) (, ), saddle point. (c) (, ) and (1, 1), both saddle points. (b) ( 1, 1), local minima, (, ), local maximum. (d) (, ), local maximum. (e) (, ), local maximum.
10 16.(7pts) Find the curvature when t = of r(t) = cos t, sin t, e t. (a) 3 (b) 1 (c) 3 (d) 1 (e) 3 17.(7pts) Which of the following is a minimum of f(x, y) = x + y subject to the constraint x + y = 1? ( ) ( ) 1 3 (a) ( 1, ) (b), (c), (d) (, 1) (e) None of these points is a minimum.
11 18.(7pts) Find the direction of fastest increase of the function f(x, y) = x y 3 at the point (, 1). (a) 4, 1 (b) 3 1,, (c) 3,, (d) 4 5, 3 5 (e) 4 5, (7pts) Find the equation of the tangent plane to the surface z = x + ln(x + y) at the point ( 1, 3, 1). (a) 3x + y z = 1 (b) x + y 3z = 1 (c) 1 + 3t, 3 + t, 1 t (d) x + 3y z = 11 (e) 3x y + z = 7
12 .(7pts) Define a transformation x(u, w) = u w, y(u, w) = w u for < u <, < w <. (x, y) Compute the Jacobian of this transformation. (u, w) Recall: d ax dx = ax ln(a). (a) u w ln(u) + w u ln(w) (b) u w w u( 1 ln(u) ln(w) ) (c) (u + w )u w 1 w u 1 (d) (u + w)u w 1 w u 1 (e) 1.(7pts) Find the volume that lies inside the sphere x + y + z = and outside the cone z = x + y. (a) 3 8 π (b) π (c) π (d) 8 π (e) π 3
13 1. Solution. div F = 3 so div F dv = 6π. Hence E F ds = F ds = F ds + F ds = 6π E T D T T is oriented the way we want and D is oriented so that the normal vector points down: the unit normal is,, 1 and the field restricted to D is x + y, y + x, since this is what we get when z =. Then x + y, y + x,,, 1 ds = ds = D Hence T D F ds = 6π. x +y 1
14 . Solution. Hence To do 1 r u = 1, u cos(θ), u sin(θ) r θ =, u sin(θ), u cos(θ) i j k r u r θ = det 1 u cos(θ) u sin(θ) u sin(θ) u cos(θ) = u 3, u cos(θ), u sin(θ) r u r θ = 4u 6 + u 4 Mass = (4x 5 + 4x 3 ) ds = (4u 5 + 4u 3 ) 4u 6 + u 4 da = T π dθ 1 u 1 θ π (4u 5 + 4u 3 ) 4u 6 + u 4 du (4u 5 +4u 3 ) 4u 6 + u 4 du use the substitution v = 4u 6 +u 4, dv = (4u 5 +4u 3 ) du. Now isn t that handy. 1 (4u 5 + 4u 3 ) 4u 6 + u 4 du = Hence the mass is π = 4 3 π v dv = v 3 5 = Solution. This question boils down to figuring out whether the given surface has C as its boundary and has orientation consistent with the orientation on C. Since the annulus has both C and the inner circle as its boundary, it is immediately out. The sphere has no boundary, so that is out. Now, for a surface to have orientation consistent with the orientation of its boundary, if you walk around the boundary in the direction of its orientation with your head pointing in the direction of the normal vector field, then the surface will be on your left. This leaves the disk with upward normals as the correct answer.
15 4. Solution. The points where a possible absolute maximum and minimum can occur are points on the two constraints such that f is in the plane formed by g and h. So f is a linear combination of g and h f ( g h ) =, scalar triple product of f, g, h is zero, or the hessian of f, g, h is zero. All those occur expect for the point ( 1, 1, 1). 5. Solution. The desired angle is given by θ = arccos where stands for the dot product. 1, 1,, 1, 3 14 ( ) 1 = arccos 8 6. Solution. The parametrization is r(x, y) = x, y, ye x, x 1, y 1. The normal vector has positive zcoordinate so the orientation is up. Hence F ( r(x, y) ) = 4y, xy, xye x i j k and ds = r x r y da = det 1 ye x 1 e x da = yex, e x, 1 da. Hence Flux = 4y, xy, xye x ye x, e x, 1 da = ( 4y 3 e x xye x + xye x ) da = x 1 y y 3 e x dy dx = 4 y x 1 y 1 1 e x dx = e x dx = e + 1
16 7. Solution. F(r(t)) = cos t, t, sin t r (t) = sin t, cos t, 1 F(r(t)) r (t) = sin t cos t + t cos t + sin t C π π F dr = π F(r(t)) r (t) dt = 1 π sin t cos t dt = u du = 1 (where u = sin t) π t cos t + sin t dt = t sin t = π Thus F dr = 1 + π. ( sin t cos t + t cos t + sin t) dt 8. Solution. x(t) dt = (sin t)i + e t k ) dt sin = t,, e t dt = sin t dt, dt, e t dt = cos t + C 1, C, e t + C 3 = ( cos t + C1 )i + C j + (e t + C 3 )k 9. Solution. The gradient is given by f = y cos(xy), x cos(xy) so D u f = f u = x + y cos(xy).
17 1. Solution. The partial derivatives are given by f x = y and f y = x. The surface area is given by S = x + y + 1 da. After converting to polar coordinates we obtain π 1 D r r + 1 dr dθ = π 1 3 ( 3 π 1) dθ = 3 ( 3 1). 11. Solution. Use the Divergence Theorem to converge this into a triple integral, x dv. This becomes 3 4 x dx dz dy = 48. E
18 1. Solution. We re looking for a vector which is parallel to a b, and it s obvious that (a b)(a b) is parallel to (a b). (because a b ) However, a+b, (a b) b and (a b) a are perpendicular to a b, and (a+b) (a b) is not a vector. 13. Solution. i j k curl F = det = e y sin(z), e x, e x cos(y) x y z e x sin(y) e y cos(z) e x
19 14. Solution. The region and the density are symmetric about the yaxis, hence the x coordinate of the center of mass is. To find the ycoordinate, we compute the moment about the xaxis. In polar coordinates, π ( π ) ( 3 ) M x = x + y y da = r(r sin(θ))r dr dθ = sin(θ) dθ r 3 dr = x +y 9 The mass is Hence, M = x +y 9 ( π) ( ) cos θ r x + y da = ȳ = π rr dr dθ = M x mass(d) = = 81 4 = 81 π π = 9 π r dr dθ = π r3 3 3 = 9π 15. Solution. We first compute f x = x + 6y, f y = 6x 6y, f xx =, f xy = 6, f yy = 6. The critical points are the solutions to the equations x + 6y = 6x 6y = The only solutions is x = y =. Thus (, ) is the only critical points. The value of the Hessian 48 and we conclude that (, ) is a saddle point
20 16. Solution. κ(t) = r (t) r (t). r (t) = sin t, cos t, e t and r (t) =, 1, 1. r (t) = r (t) 3 i j k cos t, sin t, e t and r (t) = 1,, 1., 1, 1 1,, 1 = det 1 1 = 1, 1, , 1, 1 = 3 and, 1, 1 = 3. κ() =. 17. Solution. Letting g(x, y) = x + y, the equations to solve are: { f = λ g g(x, y) = 1 which, written out are: x = λx 4y = λy x + y = 1 The first equation says that either x = or λ = 1. If x =, then the third equation gives that y = ±1; so (, 1) and (, 1) are possible extrema. If λ = 1, then the second equation gives that y = which, by the third equation, gives that x = ±1. This gives (1, ) and ( ) ( ) ( 1, ) as two more possible extrema. This eliminates (, ),, 1 3, and, as possible answers. Plugging in the possible extrema, we find that f(1, ) = f( 1, ) = 1 f(, 1) = f(, 1) = and so (1, ) and ( 1, ) are the minima of f subject to x + y = 1. Thus ( 1, ) is the correct answer.
21 18. Solution. The gradient is f = x, 3y which at (, 1) is 4, 3. The direction is 4 5, x + y, Solution. Let f(x, y, z) = x + ln x + y z. Then f = x + y, 1 so that f( 1, 3, 1) = 3, 1, 1. Since f( 1, 3, 1) defines a normal vector to the surface, an equation of the tangent plane is given by or 3x + y z = 1. 3, 1, 1 x + 1, y 3, z + 1 =. Solution. x x u w u w (x, y) (u, w) = det u w y y = det u w w u w u = det w u w 1 u w ln(u) w u ln(w) = u w u w u w w u u w w u ln(u) ln(w) = u w w u( 1 ln(u) ln(w) )
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