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The top of an object is bounded by [tex] z=4-4(x^2+y^2)[/tex] and the bottom is bounded by [tex]z=(x^2+y^2)^2 -1 [/tex], find the volume.

So I converted to polar coordinates such that

top: [tex]z=4-4r^2[/tex]

bot: [tex]z=r^2 - 1[/tex]

Then I integrated the first quadrant and multiplied by 4, with the bounds of dzdrdø such that the volume was:

[tex]V= 4[ \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=0}^{z=4-4r^2} r dz dr d \theta + \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=r^2-1}^{z=0} r dz dr d\theta ][/tex]

The answer I got was 7pi/2, but the answer from the book is 8pi/3. I think I have done all the arithematic right, but did I set it up right?

So I converted to polar coordinates such that

top: [tex]z=4-4r^2[/tex]

bot: [tex]z=r^2 - 1[/tex]

Then I integrated the first quadrant and multiplied by 4, with the bounds of dzdrdø such that the volume was:

[tex]V= 4[ \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=0}^{z=4-4r^2} r dz dr d \theta + \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=r^2-1}^{z=0} r dz dr d\theta ][/tex]

The answer I got was 7pi/2, but the answer from the book is 8pi/3. I think I have done all the arithematic right, but did I set it up right?

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