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\( \iiint_E y^2 z^2 dV \), where \(E\) is bounded by the paraboloid \(x = 1 - y^2 - z^2\) and the plane \(x = 0\)

          \( \iiint_E y^2 z^2 dV \), where \(E\) is bounded by the paraboloid \(x = 1 - y^2 - z^2\) and the plane \(x = 0\)
        
y^2 z^2 dV, where E is bounded by the paraboloid x = 1 - y^2 - z^2 and the plane x = 0

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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111. ∫y^2 dV, where E is bounded by the paraboloid r=1 and the plane z=0.
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Transcript

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00:01 Alright, so we want to go ahead and find the volume bound in between these two surfaces.
00:06 Z is equal to 2 minus x squared minus y squared and z is equal to 1.
00:12 In our case, we're going to go ahead and write our 3d space and draw these out.
00:18 So we know z equals 1 is just a plane of a plane with height at z is equal to 1.
00:26 And our paraboloid has a pika 2 and it's pointed down.
00:31 So we have this green region with the red as its bottom base for the region we have described.
00:42 Now we know that our volume is described as the double one over r, which is the region containing both planes or the intersection of the plane or other surfaces.
00:56 In this case, we'll call the z1 and z2, and it's really the top function minus the bottom function, or in this case z1 minus z2, and all with respect to, all d .a, or respect to a, or the area element.
01:12 Now we want to find this intersection region, which is just where z1 is equal to z2...
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