Question

Triple integration with spherical coordinates: ∫∫∫E e^(sqrt[x^2+y^2+z^2]); E is bounded by the xz-plane and the hemispheres: y=sqrt[9-x^2-z^2] and y=sqrt[16-x^2-z^2]

Answer 1

We first need to figure out what E looks like.

The equations \[y=\sqrt{9-x ^{2}-z ^{2}} \]
and \[y=\sqrt{16-x ^{2}-z ^{2}}\]

describe the right halves of spheres with radii 3 and 4, respectively.

The xz-plane gives us the left bound of E.

So E is the region between the spheres of radii 3 and 4 and to the right of the xz-plane.

Using spherical coordinates, the bounds on E are

\[0\le \theta \le \pi\]
\[0\le \phi \le \pi\]
\[3\le \rho \le4\]

and knowing that

\[\sqrt{x ^{2}+y ^{2}+z ^{2}}=\rho\]

We write

\[\int\limits_{}^{}\int\limits_{}^{}\int\limits_{}^{}E dV=\int\limits_{3}^{4}\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}e ^{\rho}*\rho ^{2}\sin \phi d \phi d \theta d \rho \]

\[=\int\limits_{3}^{4}\int\limits_{0}^{\pi}2\rho ^{2}e ^{\rho}d \theta d \rho\]

\[=\int\limits_{3}^{4}2\pi \rho ^{2}e ^{\rho}d \rho\]

At this point you can just integrate by parts....