Jim Lambers MAT 280 Fall Semester Practice Final Exam Solution

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1 Jim Lambers MAT 8 Fall emester 6-7 Practice Final Exam olution. Use Lagrange multipliers to find the point on the circle x + 4 closest to the point (, 5). olution We have f(x, ) (x ) + ( 5), the square of the distance from (x, ) to (, 5), as the objective function that is to be minimized. The constraints are given b g(x, ) x + 4. We then have f x λg x, f λg, where λ is the Lagrange multiplier. This gives (x ) λx, ( 5) λ, x + 4. From the first two equations, we obtain two distinct expressions for λ. Equating them ields x x 5 which simplifies to 5x. ubstituting this into the constraint ields 6x 4, or x ±.9. We then have the points (x, ) ±(.9,.966). ubstituting these into f(x, ) gives the minimizer (.9,.966).. Evaluate the iterated integral olution We have π π sin x dz d dx sin x dz d dx. π sin x dx cos x π ( ( ) ( )) u/. z d dz d d u / du, u, du d. Evaluate + x da, where is the region bounded b x, and x.

2 olution We have + x da 4 x + x d dx + x x + x 4 ln u ln. 4 x x + x dx d dx dx u du, u + x, du x dx 4. Evaluate (x + ) / da, where is the region in the first quadrant bounded b the lines and x, and the circle x + 9. olution Let f(x, ) (x + ) /. onverting to polar coordinates, we obtain (x + ) / da π/ π/ π/ π/ π r 5 5 8π 5. f(r cos θ, r sin θ)r dr (r ) / r dr r 4 dr r 4 dr The upper limit of θ, π/, is obtained using the fact that the slope of the line x,, is equal to tan π. 5. Evaluate H z x + + z dv, where H is the solid hemisphere that lies above the x-plane and has center at the origin with radius.

3 olution Using spherical coordinates, we obtain H z x + + z dv π/ π π u π 9. π/ u du ρ6 6 6 (ρ cos φ) ρ(ρ sin φ) dρ dφ cos φ sin φ dφ ρ 5 dρ, u cos φ, du sin φ dφ 6. Find the volume of the solid bounded b the clinder x + 4 and the planes z and + z. olution Using clindrical coordinates, we obtain 7. Use the transformation to evaluate V r sin θ ( r r dz dr r( r sin θ) dr ) r sin θ 6 8 sin θ (6θ + 8 ) π cos θ π. u x, R v x + x x + da, where R is the square with vertices (, ), (, ), (, ), and (, ). olution olving the above equations x and, we obtain x (u + v), (v u). This ields (x, ) (u, v) x u v u x v ( ).

4 ubstituting the vertices of the square into the change of variable ields the transformed vertices in uv-space: (, ), (, ), (, 4), and (, 4). This ields the integral R x x + da 4 4 u v v dv ln v 4 (x, ) (u, v) du dv u u du (ln 4 ln )( ) ln. 8. Evaluate the line integral x dx + e d + xz dz, where is given b r(t) t 4, t, t, t. olution From r (t) 4t, t, t, we obtain x dx + e d + xz dz 9. how that the vector field x(t)(t) dx dt + e(t) d dt + x(t)z(z)dz dt dt t (4t ) + e t (t) + t 7 (t ) dt 4t 6 + te t + t 9 dt ( ) 4t et + t ( e + ). F(x,, z) e, xe + e z, e z is conservative, and use this fact to evaluate F dr where is the line segment from (,, ) to (4,, ). olution A vector field F P, Q, R is conservative if From R Q z, P z R x, Q x P. R e z Q z, P z R x, Q x e P, we find that F is in fact conservative. To evaluate the line integral efficientl, we need to find a function f such that f F. To that end, we obtain f(x,, z) P dx xe + g(, z).

5 The requirement that f Q ields the equation g (, z) Q(x,, z) (xe ) e z. olving the equation for g ields g(, z) e z + h(z). The requirement that f z R ields the equation h (z) R(x,, z) (xe + e z ) z. It follows that h(z) K where K is an arbitrar constant, and therefore f(x,, z) xe + e z + K. From the Fundamental Theorem of Line Integrals, we obtain F dr f dr f(4,, ) f(,, ) (4e + e ) (e + e ).. Use Green s Theorem to evaluate x dx x d, where is the circle x + 4 with counterclockwise orientation. olution Let { (x, ) x + 4 } be the interior of. Then, b Green s Theorem, x dx x d ( x ) x (x ) da x da. onverting to polar coordinates, we obtain x dx x d π 8π. ( r4 4 ( r )r dr ) r dr. If f(x,, z) and g(x,, z) are twice differentiable functions, show that (fg) f g + g f + f g, where f ( f).

6 olution We have (fg) ( (fg)) (fg) x, (fg), (fg) z f x g + fg x, f g + fg, f z g + fg z (f x g + fg x ) x + (f g + fg ) + (f z g + fg z ) z f xx g + f x g x + fg xx + f g + f g + fg + f zz g + f z g z + fg zz f(g xx + g + g zz ) + g(f xx + f + f zz ) + (f x g x + f g + f z g z ) f ( g) + g ( f) + f g f g + g f + f g. In the last steps, we have used the fact that ( f) f xx + f + f xx.. Evaluate the surface integral (x z + z) d, where is the part of the plane z 4 + x + that lies inside the clinder x + 4. olution We use the parametric equations for which x u cos v, u sin v, z 4 + u cos v + u sin v, u, v π, r u cos v, sin v, cos v + sin v, r v u sin v, u cos v, u sin v + u cos v, This ields (x z + z) d r u r v u, u, u, r u r v u. [(u cos v) + (u sin v) ](4 + u cos v + u sin v) u du dv 4u + u 4 cos v + u 4 sin v du dv [u 4 + u5 (cos v + sin v) (cos v + sin v) dv 5 ] dv 6 (π) + 5 π. (sin v cos v) π. Evaluate the surface integral F d, where F(x,, z) xz,, x

7 and is the sphere x + + z 4 with outward orientation. olution We use spherical coordinates which ields x sin φ cos θ, sin φ sin θ, z cos φ, θ π, φ π, r φ cos φ cos θ, cos φ sin θ, sin φ, r θ sin φ sin θ, sin φ cos θ,, r φ r θ 4 sin φ sin φ cos θ, sin φ sin θ, cos φ, which has outward orientation since 4 sin φ for φ π. We then have F d 8 6 π π 4 8 sin φ cos θ cos φ, sin θ, cos θ 4 sin φ sin φ cos θ, sin φ sin θ, cos φ dφ sin φ( cos θ sin φ cos φ sin θ sin φ + cos θ cos φ) dφ cos θ cos θ π π + cos θ 8 cos θ 6π sin4 π φ 4 + 6π ( 6π cos φ cos φ ( 6π + ) 64π. π sin φ cos φ dφ 6 sin θ sin φ dφ + sin φ cos φ dφ π ) π u du Here, we have used the trigonometric identities (u sin φ, du cos φ dφ) ( cos φ) sin φ dφ + 4 sin θ π π ( v ) dv (v cos φ, dv sin φ dφ) sin φ cos φ dφ cos θ + cos θ, sin θ cos θ. 4. Use tokes Theorem to evaluate F dr, where F(x,, z) x, z, xz and is the triangle with vertices (,, ), (,, ) and (,, ), oriented counterclockwise as viewed from above. Hint: to obtain an equation for the surface enclosed b, compute the equation of a plane containing the vertices. olution Using the given vertices, it can be determined that is contained within the plane x + + z. We therefore describe using the parametric equations x u, v, z u v, u, v u,

8 which ields and the normal vector r u,,, r v,,, r u r v,,, which is consistent with the counterclockwise orientation of (that is, when traversing such that this normal vector, which points upward, is visible, then the region is on the left. Appling tokes Theorem ields From We then have and thus We conclude that F dr u curl F(u, v) (r u r v ) dv du. curl F R Q z, P z R x, Q x P (xz) (z) z, (x) z (xz) x, (z) x (x), z, x. curl F (r u r v ), z, x,, (x + + z), curl F(u, v) (r u r v ) (u + v + u v). F dr u dv du A(), where is the triangle { (u, v) u, v u }. This triangle has base and height, which ields F dr. 5. Use the ivergence Theorem to evaluate the surface integral F d, where F(x,, z) x,, z and is the surface of the solid E bounded b the clinder x + and the planes z and z. olution Using clindrical coordinates, we obtain F d div F dv E [(x ) x + ( ) + (z ) z ] dv E (x + + z ) dv E (r + z )r dz dr (r z + r z ) dr

9 ( ) r 4 6π + 4r π. r + 8 r dr

Jim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt

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