Question: Given $S$ as the solid that is outside the cylinder $x^{2}+y^{2}=1,$ inside the sphere $x^{2}+y^{2}+z^{2}=4,$ and above the $x y$ -plane. Set-up an iterated triple integral in spherical coordinates that is equal to $\iiint\_{G} f(x, y, z) d V$ where $f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$.

How I approached this problem is that since

$f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$ and we know that from rectangular to spherical we have $\rho=\sqrt{x^2+y^2+z^2}$ and $x=\rho sin\phi cos\theta$

I know have $\iiint\_{G} f(x, y, z) d V = \iiint\_{G} \rho sin\phi cos\theta\*\rho^2 sin\phi d\rho d\theta d\phi $

I've also determined the bounds for $\theta = 0$,$ 2\pi$ as well as for $\phi = 0$,$\pi/2$

My problem is determining the bounds for $\rho$ as the inner figure is a cylinder. My guess is that the upper bound of $\rho=2$ but I can't determine the lower bound.

Any help would be appreciated. Thanks.

Answer: 
Hint: For each $n$ find the unique $x\_n>0$ such that

$$\tag 1 ae^{-anx\_n} = be^{-bnx\_n}.$$

On $(x\_n,\infty)$ the left side of $(1)$ is greater than the right side. Thus

$$\int\_0^\infty|f\_n| > \int\_{x\_n}^\infty|f\_n| =\int\_{x\_n}^\infty f\_n.$$
Let $n\in \mathbb{N}$ fixed, now note that $$a\exp(-ax)=b\exp(-bx) \iff x={{\ln\left(\frac{b}{a}\right)}\over {b-a}}$$Call this positive number $t$, then for $x\in [0,t/n]$ we have that $$b\exp(-bnx)\ge a\exp(-anx)$$
Therefore, e have \begin{eqnarray\*}
\int\_0^\infty |a\exp(-anx)-b\exp(-bnx)|dx &\ge& \int\_0^{t/n} (b\exp(-bnx)-a\exp(-anx))dx\\
&=&{1\over n} \int\_0^t(be^{-bt}-ae^{-at})dt
\end{eqnarray\*} This integral is constant and positive, and the result follows