Use a triple integral in cylindrical coordinates to find the volume of of the gi

Question

Use a triple integral in cylindrical coordinates to find the volume of of the given solid G that is bounded above by the hemisphere z= sqrt(25-x^2-y^2), below by the xy-plane, and laterally by the cylinder x^2+y^2=9

Explanation / Answer

In cylindrical coordinates? Yikes! Let's see what happens... Due to symmetry, it suffices to find the volume in the first octant, and multiply that by 2³ = 8. For this region in the first octant, it's easy to see that 0 = z = c v[1 - (x/a)² + (y/b)²] This projects into the xy-plane as (x/a)² + (y/b)² = 1 with x,y = 0. ==> 0 = y = b v[1 - (x/a)²] with 0 = x = a. Using the standard cylindrical coordinates x = r cos ?, y = r sin ?, z = z will not cancel the 'ill effects' of a and b. (The resulting integration will be nasty.) Modify the cylindrical coordinates as follows: x = ar cos ?, y = br sin ?, z = z. Then, the Jacobian ?(x,y,z)/?(r,?,z) = |.a cos ?....b sin ?....0| |-ar sin ?...br cos ?...0| = abr |.....0...........0..........1| Better yet, the eighth of an ellipsoidal region transforms to 0 = z = c v(1 - r²), 0 = r = 1 and 0 = ? = p/2. Finally, V = 8 ???(1st octant) dV = 8 ?(? = 0 to p/2) ?(r = 0 to 1) ?(z = 0 to cv(1 - r²)) 1 (abr dz dr d?) = 8ab ?(? = 0 to p/2) ?(r = 0 to 1) r [cv(1 - r²) - 0] dr d? = 8abc [?(? = 0 to p/2) d?] * [?(r = 0 to 1) rv(1 - r²) dr] = 8abc * (p/2) * [(-1/3)(1 - r²)^(3/2) {for r = 0 to 1} = 4pabc/3.

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