asked 2014-07-23 12:18:56 -0600
I'm having problems with #5 on the homework for Spherical and Cylindrical Problems: "Figure 1 shows a 3D domain stuck between $z=x^2+y^2$ and $z=8-(x^2+y^2)$. Find the mass of the corresponding object."
so far all I have come up with is $$\int\int\int_{r^2}^{8-r^2} (r^2) dzrdrd\Theta$$
Am I on the right track at least? If so, how do I find the domain of r and $\Theta$?
answered 2014-07-23 16:18:06 -0600
I feel like it is a great start. Now if you were to set the two equations, $8-r^2$ and $r^2$, equal to each other you'll get $2r^2=8$ or $r^2=4$, which is $r=2$. So your radial boundary will go from 0 to 2 and seeing as it is a non-restricted 3d domain, $\theta$ will travel from $0$ to $2\pi$.
Ending up with: $$\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2}r^2rdzdrd\theta$$
Edit: Sorry, mistake fixed. Thanks SMS.
answered 2014-07-23 16:58:11 -0600
I agree. But I think there should be an r in there to account for the larger volume at greater radii like we said in class.
$$\int_{0}^{2 \pi}\int_{0}^{2}\int_{r^{2}}^{8-r^{2}}r^{2} rdzdrd \theta$$ $$=\int_{0}^{2 \pi}\int_{0}^{2}\int_{r^{2}}^{8-r^{2}}r^{3} dzdrd \theta$$ $$=\int_{0}^{2 \pi}\int_{0}^{2}8r^{3}-2r^{5}drd\theta$$ $$=\int_{0}^{2 \pi}\frac{32}{3}d\theta$$ $$=\frac{64 \pi}{3}$$ Let me know if you get a different answer.
Asked: 2014-07-23 12:18:56 -0600
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Last updated: Jul 23 '14
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