15_5_10 - y 2 + z 2 = a 2 is S = 8 ZZ T a dx dy √ a 2-x 2...

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10. One-eighth of the required area lies in the first octant, above the triangle T with vertices ( 0 , 0 , 0 ) , ( a , 0 , 0 ) and ( a , a , 0 ) . (See the figure.) The surface x 2 + z 2 = a 2 has normal n = x i + z k , so an area element on it can be written dS = | n | | n k | dx dy = a z dx dy = a dx dy a 2 - x 2 . The area of the part of that cylinder lying inside the cylinder
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Unformatted text preview: y 2 + z 2 = a 2 is S = 8 ZZ T a dx dy √ a 2-x 2 = 8 a Z a dx √ a 2-x 2 Z x dy = 8 a Z a x dx √ a 2-x 2 = -8 a p a 2-x 2 ± ± ± a = 8 a 2 sq. units. x y z y 2 + z 2 = a 2 x 2 + z 2 = a 2 T ( a , a , ) Fig. 5-10...
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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