Find the integral over the cone

**wopashui** · Jan 31st 2011, 02:58 PM

Evaluate the integral of $5(x^2+y^2+z^2)$ over the region D above the cone $3z^2=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=a^2$ .
(D looks like an ice-cream cone)

obviously, we may use the sphercical polar form to solve this, but I have trouble converting x, y, z into the polar from such as $x=ecos\theta$ $sin\varphi$ , $y=esin\theta$ $sin\varphi$ , $z =ecos\varphi$

**HallsofIvy** · Jan 31st 2011, 04:01 PM

I presume you mean $\rho$ rather than "e"!

Certainly you should have no trouble putting $5(x^2+ y^2+z^2)$ and $x^2+ y^2+ z^2= a^2$ into spherical coordinates ( $x^2+ y^2+ z^2= a^2$ is a sphere with radius a. So I assume the problem is $3z^2= x^2+ y^2$ . It's not really hard to put the formulas for x, y, and z, in terms of $\rho$ , $\theta$ , and $\phi$ and reduce but I think I would be inclined to add a $z^2$ to each side to get $4z^2= x^2+ y^2+ z^2$ . Now the left side is $4z^2= 4(\rho cos(\phi))^2= 4\rho^2cos^2(\phi)$ and the right side is, as I said before, very easy.

**wopashui** · Jan 31st 2011, 06:12 PM

Originally Posted by HallsofIvy

I presume you mean $\rho$ rather than "e"!

Certainly you should have no trouble putting $5(x^2+ y^2+z^2)$ and $x^2+ y^2+ z^2= a^2$ into spherical coordinates ( $x^2+ y^2+ z^2= a^2$ is a sphere with radius a. So I assume the problem is $3z^2= x^2+ y^2$ . It's not really hard to put the formulas for x, y, and z, in terms of $\rho$ , $\theta$ , and $\phi$ and reduce but I think I would be inclined to add a $z^2$ to each side to get $4z^2= x^2+ y^2+ z^2$ . Now the left side is $4z^2= 4(\rho cos(\phi))^2= 4\rho^2cos^2(\phi)$ and the right side is, as I said before, very easy.

thanks for the help, so the integral will be $\int _\,^\!$ $\int _\,^\!$ $\int _0\,^a\!$ $\rho^4$ $sin\varphi$ $d\rho$ $d\theta$ $d\varphi$

but im still not sure what the intrand of $d\theta$ $d\varphi$ would be, any hints?

would $d\varphi$ be from 0 to a/2?

**wopashui** · Feb 1st 2011, 09:17 PM

Originally Posted by wopashui

thanks for the help, so the integral will be $\int _\,^\!$ $\int _\,^\!$ $\int _0\,^a\!$ $\rho^4$ $sin\varphi$ $d\rho$ $d\theta$ $d\varphi$

but im still not sure what the intrand of $d\theta$ $d\varphi$ would be, any hints?

would $d\varphi$ be from 0 to a/2?

actually, it would be $\int _\,^\!$ $\int _\,^\!$ $\int _0\,^a\!$ $5$ $\rho^4$ $sin\varphi$ $d\rho$ $d\theta$ $d\varphi$

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