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Thread: Find the integral over the cone

  1. #1
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    Find the integral over the cone

    Evaluate the integral of $\displaystyle 5(x^2+y^2+z^2)$ over the region D above the cone $\displaystyle 3z^2=x^2+y^2$ and inside the sphere $\displaystyle x^2+y^2+z^2=a^2$.
    (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 $\displaystyle x=ecos\theta$$\displaystyle sin\varphi$, $\displaystyle y=esin\theta$$\displaystyle sin\varphi$, $\displaystyle z =ecos\varphi$
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  2. #2
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    I presume you mean $\displaystyle \rho$ rather than "e"!

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

    Certainly you should have no trouble putting $\displaystyle 5(x^2+ y^2+z^2)$ and $\displaystyle x^2+ y^2+ z^2= a^2$ into spherical coordinates ($\displaystyle x^2+ y^2+ z^2= a^2$ is a sphere with radius a. So I assume the problem is $\displaystyle 3z^2= x^2+ y^2$. It's not really hard to put the formulas for x, y, and z, in terms of $\displaystyle \rho$, $\displaystyle \theta$, and $\displaystyle \phi$ and reduce but I think I would be inclined to add a $\displaystyle z^2$ to each side to get $\displaystyle 4z^2= x^2+ y^2+ z^2$. Now the left side is $\displaystyle 4z^2= 4(\rho cos(\phi))^2= 4\rho^2cos^2(\phi)$ and the right side is, as I said before, very easy.
    thanks for the help, so the integral will be $\displaystyle \int _\,^\!$ $\displaystyle \int _\,^\! $ $\displaystyle \int _0\,^a\! $ $\displaystyle \rho^4$$\displaystyle sin\varphi$$\displaystyle d\rho$$\displaystyle d\theta$$\displaystyle d\varphi$

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

    would $\displaystyle d\varphi$ be from 0 to a/2?
    Last edited by wopashui; Jan 31st 2011 at 06:51 PM.
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  4. #4
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    Quote Originally Posted by wopashui View Post
    thanks for the help, so the integral will be $\displaystyle \int _\,^\!$ $\displaystyle \int _\,^\! $ $\displaystyle \int _0\,^a\! $ $\displaystyle \rho^4$$\displaystyle sin\varphi$$\displaystyle d\rho$$\displaystyle d\theta$$\displaystyle d\varphi$

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

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

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