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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 / 21) . 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 ) = hsin t, cos t i and hence Z C F · d r =Z 2 π h sin t,cos t i · hsin t, cos t i dt = Z 2 π dt = 2 π. 2...
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 Spring '08
 Skrypka
 Math, Vector Calculus, Sin, Cos, Line integral, Stokes' theorem

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