1. ## improper double integral

Compute this integral.
I've tried using polar coordinates (it seems the obvious thing to do) but the domain pretty much messed everthing up. since D is the upper right square [0,1]X[0,1].
I got 2 possible boundaries to choose from:
1) 0<=R<=sqrt(2) , 0<=Theta<=sin(1/R)
2) 0<=R<=1/sin(Theta) , 0<=Theta<=pie/2

I couldn't integrate either of them.. your assistance is welcome :-)

2. Originally Posted by zokomoko
Compute this integral.
I've tried using polar coordinates (it seems the obvious thing to do) but the domain pretty much messed everthing up. since D is the upper right square [0,1]X[0,1].
I got 2 possible boundaries to choose from:
1) 0<=R<=sqrt(2) , 0<=Theta<=sin(1/R)
2) 0<=R<=1/sin(Theta) , 0<=Theta<=pie/2

I couldn't integrate either of them.. your assistance is welcome :-)
If you switch to polar coordinates won't you just have

$\int_{\theta = 0}^{\pi/4} \int_{r = 0}^{r = 1/\cos \theta} dr \, d \theta + \int^{\pi/2}_{\theta = \pi/4} \int_{r = 0}^{r = 1/\sin \theta} dr \, d \theta$

and the resulting integrals in $\theta$ are not nasty.

3. ## thanks

I got stuck with another problem, it says "compute the following integral. hint: use the rotational transformation to get new coordinates (u,v) so the u = ..."

I didn't understand how the substitution they suggested is a rotational transformation, and consequently what v(x,y) should be.

4. Originally Posted by zokomoko

I've tried using polar coordinates
To do that, split the original square into two triangles:

\begin{aligned}
\int_{0}^{1}{\int_{0}^{1}{\frac{dx\,dy}{\sqrt{x^{2 }+y^{2}}}}}&=2\int_{0}^{1}{\int_{0}^{x}{\frac{dy\, dx}{\sqrt{x^{2}+y^{2}}}}} \\
& =2\int_{0}^{\frac{\pi }{4}}{\int_{0}^{\sec \varphi }{dr}\,d\varphi }=2\int_{0}^{\frac{\pi }{4}}{\sec \varphi \,d\varphi } \\
& =2\ln \left| \sec \varphi +\tan \varphi \right|\bigg|_{0}^{\frac{\pi }{4}}=2\ln \left( \sqrt{2}+1 \right).
\end{aligned}