How do you convert the following triple integral into spherical coordinates?
I am having some trouble converting the following triple integral into spherical coordinates:$$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{0}^{\sqrt{4-x^2-y^2}}e^{-\left ( x^2+y^2+z^2 \right )^\frac{3}{2}}dzdydx$$
I can convert the equation to spherical coordinates as follows (considering I did this accurately):$$e^{-\left ( \rho \right )^\frac{3}{2}}\rho^{2}sin\phi $$and then, of course, x is $\rho$, y is $\theta$, and z is $\phi$. I am unsure how to compute the new bounds for my integral, however. Any help is appreciated, and if I made any mistakes in my conversion thus far, any corrections are also appreciated!
$\endgroup$1 Answer
$\begingroup$From the innermost integral, you can notice that this is the top half of a sphere with radius $2$ (my tip on visualizing bounds for multiple integrals is to start at the innermost bounds and work your way out). From this, you can get that $0\leq\rho\leq 2$, $0\leq\theta\leq 2\pi$, and $0\leq\phi\leq \pi/2$. Also, $x^2+y^2+z^2=\rho^2$, so your integrand would be $e^{-\rho^3}\cdot\rho^2\sin(\phi)$.
$\endgroup$
