# Math Help - [SOLVED] Another improper integral

1. ## [SOLVED] Another improper integral

$\int \int_D \frac{dxdy}{\sqrt{xy}}$ where $D = \left\{x+y\leq1, x> 0, y> 0\right\}$

$\int_b^1 \left(\int_b^{1-x} \frac{dy}{\sqrt{xy}}\right)dx = 2\int_b^1 \frac{1}{x}(\sqrt{x-x^2} -\sqrt{bx})dx$

I can't solve the last integral, and the answer I get from Mathematica does not work for these limits. So I suspect the above isn't correct.

EDIT: Here's another tricky improper integral:

$\int \int \int_D \frac{dxdydz}{\sqrt{x^2+y^2+z^2}}$ where $\sqrt{x^2+y^2} < z < 1$

EDIT: And another one:

$\int \int_{R^2} \frac{xdxdy}{(x^2+y^2)^2 +1}$

2. Are you sure that isn't dydx instead of dxdy?. The way it is you have a dependent limit.

3. Originally Posted by Spec
I can't solve the last integral, and the answer I get from Mathematica does not work for these limits.
Hey Spec. This is what I get with Mathematica 6: What's wrong with it?

$\text{If}\left[\left(\text{Re}\left[\frac{b}{1-b}\right]\geq 0\&\&\frac{b}{-1+b}\neq 0\right)\left\|\text{Re}\left[\frac{b}{1-b}\right]\leq -1\right\|\text{Im}\left[\frac{b}{1-b}\right]\neq 0]\right]$

$
\frac{-\frac{4}{\sqrt{\frac{1}{b}}}+\left(2 \left(\sqrt{-1+b} \sqrt{b} \left(-1+2 \sqrt{-\frac{b}{-1+b}}\right)+\text{Log}\left[\sqrt{-1+b}+
\sqrt{b}\right]\right)\right)}{\left(\sqrt{-1+b} \sqrt{\frac{1}{b}} \sqrt{\frac{b}{1-b}}\right)}$

Also, if you want to cut-and-paste Mathematica as latex code, select the code in Mathmatica, then choose Edit/Copy as/Latex (with some minor limitations). For example, I didn't type in any of that code above although I did have to re-format it slightly because of some limitations on latex length in here.

4. The integral (and answer) looks way too complicated, and we're supposed to be able to solve all these problems without using any tools.

5. Originally Posted by Spec
$\int \int_D \frac{dxdy}{\sqrt{xy}}$ where $D = \left\{x+y\leq1, x> 0, y> 0\right\}$

$=\int_0^1 \frac{2}{\sqrt{x}} [ \sqrt{y}]_0^{1-x} \ dx=2\int_0^1 \sqrt{\frac{1-x}{x}} \ dx=4\int_0^{\frac{\pi}{2}} \cos^2 \theta \ d \theta = \pi. \ \ \ \square$

in the last integral we put $x=\sin^2 \theta.$

$\int \int \int_D \frac{dxdydz}{\sqrt{x^2+y^2+z^2}}$ where $\sqrt{x^2+y^2} < z < 1$

$=\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^{\sec \phi} \rho \sin \phi \ d \rho \ d \phi \ d \theta=\pi \int_0^{\frac{\pi}{4}} \sec^2 \phi \ \sin \phi \ d \phi = \pi [\sec \phi]_0^{\frac{\pi}{4}}=\pi(\sqrt{2} \ - \ 1). \ \square$

note that in spherical coordinates $z=\sqrt{x^2+y^2}$ becomes $\phi = \frac{\pi}{4}$ and $z=1$ becomes $\rho=\sec \phi.$

$\int \int_{R^2} \frac{xdxdy}{(x^2+y^2)^2 +1}$

$= \int_0^{\infty} \int_0^{2 \pi} \frac{r^2 \cos \theta}{r^4 + 1} \ d \theta \ dr = \int_0^{\infty} \frac{r^2}{r^4 + 1} [\sin \theta]_0^{2 \pi} \ dr=0. \ \ \ \square$

note that $\int_0^{\infty} \frac{r^2}{r^4 + 1} \ dr$ is convergent.

6. Originally Posted by NonCommAlg
$= \int_0^{\infty} \int_0^{2 \pi} \frac{r^2 \cos \theta}{r^4 + 1} \ d \theta \ dr = \int_0^{\infty} \frac{r^2}{r^4 + 1} [\sin \theta]_0^{2 \pi} \ dr=0. \ \ \ \square$

note that $\int_0^{\infty} \frac{r^2}{r^4 + 1} \ dr$ is convergent.
Does that matter though? Since your multiplying it with zero either way. I actually came that far myself, but struggled with the calculation of $\int_0^{\infty} \frac{r^2dr}{r^4 + 1}$ (I tried to integrate it before $d\theta$, and thus didn't notice that the $d\theta$ integral was zero).