15_5_12 - 2 is dS 2 = | n 2 | | n 2 • j | dx dz = b dx dz...

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12. We want to find A 1 , the area of that part of the cylinder x 2 + z 2 = a 2 inside the cylinder y 2 + z 2 = b 2 , and A 2 , the area of that part of y 2 + z 2 = b 2 inside x 2 + z 2 = a 2 . We have A 1 = 8 × ( area of 1 ), A 2 = 8 × ( area of 2 ), where 1 and 2 are the parts of these surfaces lying in the first octant, as shown in the figure. A normal to 1 is n 1 = x i + z k , and the area element on 1 is dS 1 = | n 1 | | n 1 i | dy dz = a dy dz a 2 - z 2 . x y z y 2 + z 2 = b 2 1 2 R 2 R 1 a b x 2 + z 2 = a 2 Fig. 5-12 A normal to 2 is n 2 = x j + z k , and the area element on
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Unformatted text preview: 2 is dS 2 = | n 2 | | n 2 • j | dx dz = b dx dz √ b 2-z 2 . Let R 1 be the region of the first quadrant of the yz-plane bounded by y 2 + z 2 = b 2 , y = 0, z = 0, and z = a . Let R 2 be the quarter-disk in the first quadrant of the xz-plane bounded by x 2 + z 2 = a 2 , x = 0, and z = 0. Then...
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