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

$$\int\_0^\infty|f\_n| > \int\_{x\_n}^\infty|f\_n| =\int\_{x\_n}^\infty f\_n.$$
find out when $f\_n>0$ or $<0$ so:
$$0=f\_n=ae^{-anx}-be^{-bnx}$$
$$a(e^{-nx})^a=b(e^{-nx})^b$$
$$\frac ab=(e^{-nx})^{b-a}$$
$$\ln a-\ln b=(b-a)(-nx)$$
so:
$$x=\frac{\ln a-\ln b}{n(a-b)}$$
is when $f\_n=0$

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$$\int\_0^\infty f\_ndx=\int\_0^\infty\left(ae^{-anx}-be^{-bnx}\right)dx=\left[\frac{e^{-anx}-e^{-bnx}}{-n}\right]\_{x=0}^\infty=0$$

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$$\int\_0^\infty|f\_n|dx=-\int\_0^{\frac{\ln a-\ln b}{n(a-b)}}(ae^{-anx}-be^{-bnx})dx+\int\_{\frac{\ln a-\ln b}{n(a-b)}}^\infty(ae^{-anx}-be^{-bnx})dx$$
Try and find the value of this then plug it into the summation