Evaluating $\int_{x=1}^{x=2}\int_{y=0}^{y=x}\frac{1}{(x^2+y^2)^{3/2}}\,\mathrm{d}y\,\mathrm{d}x$ using polar coordinates

Question

$$\int_{x=1}^{x=2}\int_{y=0}^{y=x}\frac{1}{(x^2+y^2)^{3/2}}\,\mathrm{d}y\,\mathrm{d}x$$

Can anyone please help me solve this question through the use of polar coordinates? I tried doing it, replaced $x^2 +y^2 =r^2$ and then plotted the region on $x-y$ coordinates (it was the region bounded by the line $y=x$ and the the lines $x=1$ and $x=2$. The problem is, since the region is not circular, how do I convert the respective limits into polar form? I thought of substituting $y=x$ by $y=r\cos(\theta)$ but if I am to integrate first wrt $r$ then my limits should not contain $r$ right?

Can anyone please help solve this?

Many thanks.

Answer 1

The angle goes from $0$ to $\frac{\pi}{4}$.

Upon fixing the angle, let's study how does the radius behave.

Let the lower limit be $r_1$ and upper limit be $r_2$,

then $\cos(\theta) = \frac1{r_1} = \frac{2}{r_2}$

That is $r_1 = \sec\theta$ and $r_2 = 2\sec\theta$.

Answer 2

Try $\theta\in [0,\frac\pi4]$ and $r\in [\frac1{\cos\theta},\frac2{cos\theta}]$, to sweep out the region...

Answer 3

The angle varies from 0 to 45 degrees as y varies from 0 to x.

For the limits of r you can use the fact that r1 cos(theta) = 1 and r2 cos(theta) = 2, for minimum and maximum values of r.