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
Best 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 $$ \int_{x^2+y^2<a^2}dx dy |x|=\int_0^a dr\ r\int_0^{2\pi}r |cos\theta|d\theta=4\int_0^a dr r^2=\frac{4}{3}a^3\ . $$