Finding $\iiint_R e^{\sqrt{(x^2+y^2+z^2)^3}} dA$ with spherical coordinates

Question

How to integrate the following expression

$$\iiint_R e^{\sqrt{(x^2+y^2+z^2)^3}} dA\,,$$

where $R$ is a sphere with radius $1$ centered in the origin?

I did

$$\iiint_R e^{\sqrt{(x^2+y^2+z^2)^3}} dA = 4\int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 e^{\sqrt{\rho^3}} \rho^2 \sin{\phi} \,d\rho \,d \phi \,d \theta$$

But my question is if it's correct to consider the first octant, as I did, and then just multiplying it by $4$ since the exponential function relies on positive $z$-axis.

Answer 1

The exponent is actually $\sqrt{(\rho^2)^3}=\rho^3$, i.e. your last $\sqrt{}$ operator is unneeded. Then$$\int_0^1\rho^2e^{\rho^3}d\rho=[e^{\rho^3}/3]_0^1=(e-1)/3.$$The rest of the integral is just a factor of the sphere's surface, $4\pi$.

The variable $\sqrt{x^2+y^2+z^2}$ is more normally denoted $r$, though, with $\rho$ preserved for a cylindrical coordinate $\sqrt{x^2+y^2}$.