# Double Integral using Polar Coordinate System

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: And I know that:

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

and

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

also

$$x^{2}+y^{2}=r^{2}$$

From the region it is clear that

$$0\leq r\leq 1$$

and

$$0\leq \theta \leq \pi$$

so I thought (and probably wrong):

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

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

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.
$$\displaystyle \int_0^{\pi} \int_0^1 \frac{r}{1+r^2} ~ dr ~ d\theta = \int_0^{\pi} \frac 12 \left[\ln(r^2 + 1)\right]_0^1 ~ d\theta$$
$$\displaystyle = \frac {\pi}{2} \cdot \ln2$$