# 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!

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 $$e^{-\rho^3}\cdot\rho^2\sin(\phi)$$.