15_5_7 - x 2 dx dy = Z 1 x p 1 + x 2 dx Z √ 1-x 2 dy = Z...

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7. On the surface with equation z = x 2 / 2 we have z /∂ x = x and z /∂ y = 0. Thus dS = p 1 + x 2 dx dy . If R is the first quadrant part of the disk x 2 + y 2 1, then the required surface integral is ZZ x dS = ZZ R x p 1 +
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Unformatted text preview: x 2 dx dy = Z 1 x p 1 + x 2 dx Z √ 1-x 2 dy = Z 1 x p 1-x 4 dx Let u = x 2 du = 2 x dx = 1 2 Z 1 p 1-u 2 du = 1 2 π 4 = π 8 ....
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