11.03 - Section 12.5: Triple integrals (concluded) Group...

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Section 12.5: Triple integrals (concluded) Group work 2: Setting up volume integrals, problem 2: “Find the volume of the part of the solid sphere x 2 + y 2 + z 2 =4 lying above the plane z =1.” The book gives –sqrt(3) sqrt(3) –sqrt(4– xx ) sqrt(4– xx ) 1 sqrt(4– xx yy ) dz dy dx as the answer, but this is wrong: the correct answer is –sqrt(3) sqrt(3) –sqrt(3– xx ) sqrt(3– xx ) 1 sqrt(4– xx yy ) dz dy dx . Note that S consists of points ( x , y , z ) that lie below the hemisphere z = sqrt(4– x 2 y 2 ) and above the plane z = 1; that is, we need 1 z sqrt(4– x 2 y 2 ). But for values of y that are greater than sqrt(3– x 2 ) (or less than –sqrt(3– x 2 )) we have sqrt(4– x 2 y 2 ) < 1, so there are no such values of z . Geometrically, what’s happening is that when ( x , y ) leaves the disk of radius sqrt(3) centered at (0,0), the two surfaces (the sphere and the plane) cross, with the plane now lying above the sphere rather than below. Is the incorrect answer given by Stewart too large, or too
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11.03 - Section 12.5: Triple integrals (concluded) Group...

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