Answer to Question #143562 in Calculus for Promise Omiponle

Question

Evaluate ∫∫∫D e^(sqrt(x^2+y^2+z^2)) dV, where D is the region enclosed by x^2+y^2+z^2= 4 within the first octant. Recall that the first octant is the region for which x >0; y >0; z >0.

Answer 1

Lets convert to spherical coordinates:

$x=\rho sin(\phi) cos(\theta)$

$y=\rho sin(\phi) sin(\theta)$

$z=x=\rho cos(\phi)$

$\rho^2 = x^2+y^2+z^2=4$

Since the region is enclosed by the sphere within the 1st octant let set the limits:

$\theta\in [\, 0, \frac{\pi}{2}]\,$ , $\phi\in [\, 0, \frac{\pi}{2}]\,$ , $\rho\in [\, 0, 2]\,$

Therefore

$I=\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^2e^\rho\rho^2sin(\phi)d\rho d\phi d\theta=[\,\int_0^\frac{\pi}{2}d\theta]\,[\,\int_0^\frac{\pi}{2}sin(\phi)d\phi]\,[\,\int_0^2e^\rho \rho^2d\rho]\,$

Applying integration by parts:

$\int_0^2e^\rho \rho^2d\rho=(\rho^2e^\rho-\int e^\rho2\rho d\rho)\vert_0^2=(\rho^2e^\rho-2(\rho e^\rho-e^\rho))\vert_0^2=2e^2-2$

$I=\frac{\pi}{2}\cdot(-cos({\phi})\vert_0^\frac{\pi}{2})\cdot(2e^2-2)=\pi e^2-\pi$

Answer: $\pi e^2-\pi$