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quiz10-soln

# quiz10-soln - 427L Quiz 10 Solution Nov 28 2007 Consider...

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427L Quiz 10 Solution Nov. 28, 2007 Consider the following 2 -form in R 3 : ω = xdy dz + y dz dx + z dx dy ( x 2 + y 2 + z 2 ) 3 / 2 , or upon letting r = p x 2 + y 2 + z 2 , ω = ( x/r 3 ) dy dz + ( y/r 3 ) dz dx + ( z/r 3 ) dx dy. Problem 1 (5 pts) . Show that ω is closed, i.e., that = 0 . Proof. Notice that = ± ∂x x r 3 ² dx dy dz + ± ∂y y r 3 ² dy dz dx + ± ∂z z r 3 ² dz dx dy = ± ∂x x r 3 + ∂y y r 3 + ∂z z r 3 ² dx dy dz. But ∂x x ( x 2 + y 2 + z 2 ) 3 / 2 = ( x 2 + y 2 + z 2 ) 3 / 2 - x · 3 2 · 2 x ( x 2 + y 2 + z 2 ) 1 / 2 ( x 2 + y 2 + z 2 ) 3 = r 2 - 3 x 2 r 5 . And likewise, ∂y y r 3 = r 2 - 3 y 2 r 5 , ∂z z r 3 = r 2 - 3 z 2 r 5 , so their sum is 3 r 2 - 3( x 2 + y 2 + z 2 ) r 5 = 0 , and thus = 0 dx dy dz = 0 as claimed. Problem 2 (5 pts) . Show that for the sphere of radius r given by S = { ( x,y,z ) | x 2 + y 2 + z 2 = r 2 } , we have ZZ S ω = 4 π. 1

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Hint: Use the fact that for a surface with unit normal n = ( n x ,n y ,n z ) , dA = n x dy dz + n y dz dx + n z dx dy, and recall that the unit normal on S is given by n = ( x/r,y/r,z/r ) . Thus you should conclude that ω = dA/r 2 , and remember that
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quiz10-soln - 427L Quiz 10 Solution Nov 28 2007 Consider...

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