# Solving double integral over disk

I would like to know if you guys could help me sort out this two question topic being double integrals The question is as follows: Evaluate the following double integral over disk $D$ , where $D$ is defined by ${x^2}+ {y^2}\leq{a^2}$.

a. $$\int_D \frac{dxdy}{\sqrt{x^2+y^2}}$$

b. $$\int_D dxdy|x|$$

for part a I have used $\int(1/a-r^2)rdrd\phi$ and obtained $=-(\pi/2)\ln(a-1)$as my final solution

for part b I also used direct substitution $\int rcos \phi r dr d \phi$ and obtained zero for my final solution . I think there's probably a better way to do it as I am not happy with this solution , probably by trigonometric substituition , can anyone please take a look at this and give me some tips?

Thanks in advance

## 1 Answer

Your first integral is $$\int_{x^2+y^2<a^2}dx dy \frac{1}{\sqrt{x^2+y^2}}\ ,$$ which in polar coordinates reads $$\int_0^a dr \underbrace{r}_{Jacobian}\times \frac{1}{r}\int_0^{2\pi}d\theta=2\pi a\ .$$ The second integral is $$\int_{x^2+y^2<a^2}dx dy |x|=\int_0^a dr\ r\int_0^{2\pi}r |cos\theta|d\theta=4\int_0^a dr r^2=\frac{4}{3}a^3\ .$$