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**HallsofIvy** The integral is $\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}}\int_{\sqrt{x^2+ y^2}}^1 dzdydx$.

Looking at the "outer integral", x goes from x= -1 to x= 1, and then, looking at the second integral, for each x, y goes from $\displaystyle y= -\sqrt{1- x}$ to $\displaystyle y= \sqrt{1- x^2}$. Those last two are the same as $\displaystyle y^2= 1- x^2$ or $\displaystyle x^2+y^2= 1$, the circle with center at (0, 0) in the xy-plane and radius 1. Finally, for every such (x, y), z goes from $\displaystyle z= \sqrt{x^2+ y^2}$, which is the part of a cone with vertex at (0, 0, 0) opening upwards, to z= 1. Are you **required** to do this in spherical coordinates? I think it would be easier in "cylindrical coordinates": [tex]\int_0^{2\pi}\int_0^1\int_r^1 r dzdrd\theta[tex] since, in cylindrical coordinates, that cone is just z= r.

If you really must do it in spherical coordinates, then $\displaystyle x= \rho cos(\theta)sin(\phi)$ and $\displaystyle y= \rho sin(\theta) sin(\phi)$ so $\displaystyle \sqrt{x^2+ y^2}= \sqrt{\rho^2cos^2(\theta)sin^2(\phi)+ \rho^2sin^2(\theta)sin^2(\phi)}= \sqrt{\rho^2\sin^2(\phi)(sin^2(\theta)+ cos^2(\theta))}= \rho\sin(\phi)$.

$\displaystyle z= \rho cos(\phi)$ so $\displaystyle z= \sqrt{x^2+ y^2}$ becomes $\displaystyle \rho cos(\phi)= \rho sin(\phi)$ and $\displaystyle cos(\phi)= sin(\phi)$ or $\displaystyle tan(\phi)= 1$ so that $\displaystyle \phi= \frac{\pi}{4}$ which, now that I think of it, should have been obvious! It is the upper limit, z= 1, that is more complicated. $\displaystyle z= \rho cos(\phi)= 1$ so $\displaystyle \rho= \frac{1}{cos(\phi)}= sec(\phi)$.

In spherical coordinates, the integral is $\displaystyle \int_{0}^{2\pi}\int_0^{\pi/4}\int_{0}^{sec(\phi)} \rho^2 sin^2(\phi)d\rho d\phi d\theta$