11.07 - Section 12.7 Spherical coordinates(concluded Recall...

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Section 12.7: Spherical coordinates (concluded) Recall the group work problem “Find the volume of the part of the solid sphere x 2 + y 2 + z 2 =4 lying above the plane z =1.” In Cartesian coordinates, the integral we have to evaluate is –sqrt(3) sqrt(3) –sqrt(3– xx ) sqrt(3– xx ) 1 sqrt(4– xx yy ) dz dy dx . (This is the one that the book got wrong; see the notes for last Thursday.) If we perform the inner integral, we get –sqrt(3) sqrt(3) –sqrt(3– xx ) sqrt(3– xx ) (sqrt(4– x 2 y 2 ) –1) dy dx , but then we’re stuck (the antiderivative of sqrt(4– x 2 y 2 ) –1 is quite nasty). In cylindrical coordinates, we get the integral … 0 sqrt(3) 0 2 π 1 sqrt(4– rr ) r dz d θ dr = 0 sqrt(3) 0 2 π r (sqrt(4– r 2 ) – 1) d θ dr = 2 π 0 sqrt(3) r (sqrt(4– r 2 ) – 1) dr = 2 π [(–1/3)(4– r 2 ) 3/2 r 2 /2]| 0 sqrt(3) = 5 π /3. (I didn’t have time to do it in spherical coordinates, but presumably this is a doable approach too.) Section 12.8: Change of variables in multiple integrals Main ideas?
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