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What would be the bounds for the triple integral over the surface of a cone, $$x^2+y^2 \leq z^2 $$ where $$0 \leq z \leq a$$

My first guess is to use polar coordinates for $x$ and $y$: $$x=rcos(\theta)$$ $$y=rsin(\theta)$$

I am unsure about what to use for $z$, and then I am unsure what the bounds of integration are for $x$ and $y$ when I go to set up my triple integral.

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We have $r^2\le z^2$, thus $0\le r\le z$. On the other hand $0\le \theta\le 2\pi$, therefore $$\int_{0}^{2\pi}\int_{0}^{a}\int_{0}^{z}r\,\mathrm{d}r\mathrm{d}z\mathrm{d}\theta=\frac 13 a^3 \pi$$

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  • $\begingroup$ If my function is $F=[x^2,y^2,z^2]$, then what would be the function I am integrating? I take the partials and get $2x+2y+2z$, so is it $2rcos(\theta)+2rsin(\theta)+2z$? $\endgroup$ – Gary Feb 1 '17 at 0:24
  • $\begingroup$ Apply Divergence theorem . en.wikipedia.org/wiki/Divergence_theorem $\endgroup$ – Behrouz Maleki Feb 1 '17 at 8:37
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$$\int_0^a\int_0^{z}\int_0^{2\pi}r\mathrm d\theta\mathrm dr\mathrm dz=\frac{1}{3}\pi a^3$$

One way to see this is consider a section of the cone at certain heigh z. Its area is $\pi z^2$. Then you simply have to integrate it between $0$ and $a$. The two inner integrals give you the area, the third, the volume.

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