{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Quiz 4 Solution - Math 3010 Quiz 4 Nov 28 2012 1(4 points...

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

View Full Document Right Arrow Icon
Math 3010 Quiz 4 Nov. 28, 2012 1. (4 points) For a surface S and ±xed vector vV R 3 , prove that 2 ii S · ˆ n dS = i ∂S ( × v r ) · d v S. where v r ( x, y, z ) = ( x, y, z ). Solution: Let = ( a, b, c ) and v F = × v r . Then × v r = v v v v v v ˆ ı ˆ ˆ k a b c x y z v v v v v v = ( bz cy, cx az, ay bx ) . v ∇ × ( × v r ) == v v v v v v ˆ ı ˆ ˆ k ∂x ∂y ∂z bz cy cx az ay bx v v v v v v = (2 a, 2 b, 2 c ) = 2 vV. By Stokes’ theorem, II S p v ∇ × v F P · ˆ n dS = I ∂S v F · dvs . Thus, 2 S · ˆ n dS = i ∂S ( × v r ) · d v 2. (a) (2 points) Find f : R 2 R such that ( y 3 + 1 , 3 xy 2 + 1) = v f , if it exists. Solution: This is a conservative vector ±eld since v F is de±ned everywhere on R 2 and v ∇ × v F = 0. To ±nd f , solve the following system of equations: ∂f ∂x = y 3 + 1 , ∂y = 3 xy 2 + 1 Solving for f in the ±rst equation leads to f ( x, y ) = xy 3 + x + g ( y ). Using this in the second equation yields g ( y ) = 1, so g ( y ) = y + any constant. So we can choose f ( x, y ) = xy 3 + x + y .
Image of page 1

Info icon This 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.

{[ 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