I started learning double integrals. I am trying to solve one problem but its not going well :).$$ \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \ln\bigl(x^2+y^2+1\bigr)\, dx\,dy $$
then I wrote :$$ -\sqrt{1-y^2} \leq x \leq \sqrt{1-y^2} \quad \text{and}\quad -1\leq y \leq1$$now let $x = \sqrt{1-y^2} \Rightarrow x^2 + y^2 = 1$ so I thought the region was a circle with radius $1 \implies-1\leq r \leq 1$.
Also I change $ \ln(x^2+y^2+1) \rightarrow \ln(r^2+1) $
sum up:$$\int_{?}^{?} \int_{-1}^{1} \ln\bigl(r^2+1\bigr) \,r\,dr\,d\theta$$
I don't know how to find values for $\theta$ if the region is circle then I think it should be$- \frac{\pi}{2}$ and $\frac{\pi}{2}$ but in this case I am getting $0$
I know it's easy question but I can't understand how it works.
Thanks in advance
$\endgroup$2 Answers
$\begingroup$Please note that in polar coordinates, $x = r \cos\theta, y = r \sin \theta$. As a standard practice in polar coordinates, you keep $r$ non-negative and you cover points in different quadrants by changing the value of $\theta \in (0, 2\pi)$
Here you have $ - \sqrt{1-y^2} \leq x \leq \sqrt{1-y^2}$ and $- 1 \leq y \leq 1$ which is the entire unit circle $x^2 + y^2 \leq 1$.
So in polar coordinates, $0 \leq r \leq 1$ and $0 \leq \theta \leq 2\pi$ and the integral becomes,
$ \displaystyle \int_0^{2\pi} \int_0^1 r \ln (1+r^2) \ dr \ d\theta$
$\endgroup$ $\begingroup$Using polar coordinates, the integral becomes:
$\displaystyle \int_{\theta = 0}^{2\pi} \int_{r = 0}^1 \ln(r^2 + 1) r dr d\theta $
$\endgroup$
