Spherical coordinates to calculate triple integral

Question

The region $D$ is bounded below by $z=2\sqrt{x^2+y^2}$ and above by $x^2+y^2+z^2=5$ My task is to calculate $$\iiint_D e^{{(x^2+y^2+z^2)}^{3/2}}dV.$$

I tried switching to spherical coordinates $$(x,y,z) \to (\rho \sin(\phi)\cos(\theta), \rho \sin(\phi)\sin(\theta), \rho \cos(\phi)).$$

From the cone-equation I get $\rho \cos(\phi) = 2 \rho \sin(\phi)$ and thus $\phi = \arctan(1/2)$, where the other limits are trivial.

I end up with $$\int_ 0^{2\pi}\int_0^{\arctan{\frac{1}{2}}} \int_0^{\sqrt{5}} e^{\rho^3}\cdot \rho^2\sin(\phi)d\rho d\phi d\theta.$$

Could someone confirm whether or not the bounds and my integral is correct here?

Answer 1

Yes, you are correct. The given integral in spherical coordinates is $$\int_ 0^{2\pi}\int_0^{\arctan{\frac{1}{2}}}\int_0^{\sqrt{5}}e^{\rho^3}\cdot \rho^2\cdot \sin(\phi)d\rho d\phi d\theta=2\pi\left(1-\frac{2}{\sqrt{5}}\right)\left(\frac{e^{\sqrt{5^3}}-1}{3}\right).$$ where we used the fact that $\cos(\arctan(\phi))=\frac{1}{\sqrt{1+\phi^2}}.$

Answer 2

The lower bound is

$$\rho\cos(\phi)=2\sqrt{\rho^2 \sin^2(\phi)\cos^2(\theta)+\rho^2 \sin^2(\phi)\sin^2(\theta)}=2\rho|\sin(\phi)|$$ or in the first quadrant

$$\tan\phi=\frac12,$$

which describes a cone, and the upper bound

$$5=\rho^2 \sin^2(\phi)\cos^2(\theta)+\rho^2 \sin^2(\phi)\sin^2(\theta)+\rho^2\cos^2(\phi)=\rho^2$$ or $$\rho=\sqrt5,$$ a sphere.

Hence the volume is symmetrical around $z$ and the bounds are indeed those of a cone with a spherical cap

$$\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\arctan1/2}\int_{\rho=0}^{\sqrt5}.$$