251-f06-finalsolutions

251-f06-finalsolutions - i| = r we have ZZ S 2 f x y z dS =...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 251.504 Solutions for the Final Examination Fall 2006 1. (a) true, (b) true, (c) true, (d) false, (e) false, (f) true, (g) true, (h) true. 2. (a) curl F = 0 , y cos( yz ) , - z cos( yz ) and div F = - sin y + e z . (b) curl G = - 1 , - 1 , - 2 y , and so curl F = - 2 , 0 , 0 and div F = 0. 3. We have F = f where f ( x, y, z ) = x 3 yz 2 + sin x + y , and so C F · d r = f ( π, - 2 , 0) - f (0 , 1 , 3) = - 3 by the Fundamental Theorem for Line Integrals. 4. We have div F = 6 z ( x 2 + y 2 + z 2 ) 2 and so by the Divergence Theorem the given surface integral is equal to E 6 z ( x 2 + y 2 + z 2 ) 2 dV = 1 0 π 2 0 2 π 0 (6 ρ 5 cos ϕ ) ρ 2 sin ϕ dθ dϕ dρ = 6 1 0 ρ 7 π 2 0 1 2 sin 2 ϕ dϕ 2 π 0 = 3 4 π. 5. By Green’s Theorem the given line integral is equal to 3 0 3 - x - 3+ x x dy dx = 3 0 (2 x 2 - 6 x ) dx = - 9 . 6. The cylindrical part S 1 of the surface can be parametrized by r ( θ, z ) = cos θ, sin θ, z where 0 θ 2 π and - 3 z 3, and since | r θ × r z | = | cos θ, sin θ, 0 | = 1 we have S 1 f ( x, y, z ) dS = 2 π 0 3 - 3 (2 + z 2 ) dz dθ = 60 π. The base S 2 can be parametrized by r ( r, θ ) = r cos θ, r sin θ, - 3 where 0 r 1 and 0 θ 2 π , and since
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: i| = r we have ZZ S 2 f ( x, y, z ) dS = Z 2 π Z 1 (2 r 2 + 9) r dr dθ = 10 π. Thus RR S f ( x, y, z ) dS = RR S 1 f ( x, y, z ) dS + RR S 2 f ( x, y, z ) dS = 70 π . 1 7. The area is equal to ZZ D q 1 + ( ∂z ∂x ) 2 + ( ∂z ∂y ) 2 dA = 2 Z 1 Z x √ 1 + x 2 dy, dx = 2 3 (2 3 / 2-1) . 8. The line integral around each of the four smooth parts of the boundary of S is zero, and so the given surface integral is equal to zero. 9. Using the parametrization r ( t ) = h cos t, sin t i for 0 ≤ t ≤ 2 π (which gives the opposite orientation) we have r ( t ) = h-sin t, cos t i and hence Z C F · d r =-Z 2 π h sin t,-cos t i · h-sin t, cos t i dt = Z 2 π dt = 2 π. 2...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern