evalute $\int_{0}^{3} \int_{0}^{x} \frac{1}{\sqrt{x^2+y^2}} dy \ dx$ by polar coordinates

Question

$$\int_{0}^{3} \int_{0}^{x} \frac{1}{\sqrt{x^2+y^2}} dy \ dx$$

After I sketch the area required, it is a right angle triangle with vertices $(0,0), (3,0), (3,3)$.

Now I have to change it to polar coordinates to solve it. So, I know that $ dA = dx \ dy = r\ dr\ d{\theta}$ by solving the jacobian.

the new integral should be something like $$ \iint_{D*}^{} 1\ dr\ d{\theta}$$ but I am having trouble determining the limits of $r$ and $\theta$. I think it should be $$0\le \theta \le \frac{\pi}{4}$$ and $$0\le r\le3.$$ Though I'm not sure about $r$.

Answer 1

The maximum value of $r$ varies with $\theta$. Namely, $r$ is the hypotenuse of the right triangle with vertices $(0,0), (3,0), \left(3, 3\tan\theta\right)$.

Therefore we can calculate the maximum value of $r$ as $\frac{3}{\cos\theta}$.

So the integral is

$$\int_0^{\frac\pi4} \int_0^{\frac3{\cos\theta}} 1\,dr\,d\theta = \int_0^{\frac\pi4}\frac{3\,d\theta}{\cos\theta} = \frac32 \ln\left(3+2\sqrt2\right)$$

Answer 2

You're fine but $\;r\;$ is wrong: on the $x$-axis, we have that $0\le r\le 3$, but on the ray $y=x$ we have $0\le r\le 3\sqrt2$ , so we get that $0\le x=r\cos t\le 3\implies 0\le r\le\cfrac 3{\cos t}=3\sec t.$ Try now to complete the answer...