# Evaluate \iiint_{E} \sqrt{( } x^{2}+y^{2}+z^{2} ) d V where E lies above the cone z=\sqrt{( }...

## Question:

Evaluate

{eq}\iiint_{E} \sqrt{ x^{2}+y^{2}+z^{2} }\, d V {/eq} where {eq}E {/eq} lies above the cone {eq}z=\sqrt{ x^{2}+y^{2} } {/eq} and between the spheres and between the spheres {eq}x^{2}+y^{2}+z^{2}=1 {/eq} and {eq}x^{2}+y^{2}+z^{2}=4 {/eq}.

## Iterated Integrals; Spherical Coordinates:

The region of integration {eq}E {/eq} is partially bounded by two spheres and the integrand depends only on the distance to the origin, so it is a good idea to try to evaluate the integral using spherical coordinates.

We recall that the main equations for spherical coordinates are:

{eq}\begin{align*} x&=\rho\cos\theta\sin\phi,\\ y&=\rho\sin\theta\sin\phi,\\ z&=\rho\cos\phi,\\ \rho^2&=x^2+y^2+z^2, \end{align*} {/eq}

and the volume element is given by {eq}\rho^2\sin\phi\,d\rho\,d\theta\,d\phi. {/eq}

## Answer and Explanation:

We observe region of integration {eq}E {/eq} is partially bounded by spheres and a cone; also the integrand depends only on the distance to the...

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