Calculate the following double integral using polar coordinates: $\iint_{D}^{}\frac{dxdy}{1+x^{2}+y^{2}}$

$D = \left \{(x,y) | 0\leq y\leq \sqrt{1-x^{2}} \right \}$

I have started solving but pretty sure I have mistakes.

The region is:

enter image description here

And I know that:

$x=r\cdot cos(\theta )$


$y=r\cdot sin(\theta )$



From the region it is clear that

$0\leq r\leq 1$


$0\leq \theta \leq \pi $

so I thought (and probably wrong):

$\iint_{}^{}\frac{1}{1+r^{2}}drd\theta =\theta \cdot arctan(\theta )$

  • 1
    $\begingroup$ The Jacobian is $r \ dr \ d\theta.$ $\endgroup$ Jan 14, 2022 at 12:22

1 Answer 1


You found the bounds correctly but you missed the Jacobian $r$. Note that $dx ~ dy = r ~ dr ~ d\theta ~$ as you change from cartesian coordinates to polar coordinates with $x = r \cos\theta, y = r \sin\theta$.

Also it is definite integral with $~0 \leq \theta \leq \pi~$. So the final answer cannot be a function of $\theta$. You made some mistake in integral too.

The correct integral is,

$ \displaystyle \int_0^{\pi} \int_0^1 \frac{r}{1+r^2} ~ dr ~ d\theta = \int_0^{\pi} \frac 12 \left[\ln(r^2 + 1)\right]_0^1 ~ d\theta$

$ \displaystyle = \frac {\pi}{2} \cdot \ln2$


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