Find the double integral by changing to polar coordinates

Question

So I have the following double integral

$$\int_{-2}^{0} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} dydx$$

If I integrate with respect to y first I get: $\frac{1}{2}(y\sqrt{x^2+y^2}+x^2\log(\sqrt{x^2+y^2}+y)$

I understand that in polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$, but am unsure how to convert the limits to polar coordinates? Thanks

Answer 1

Your integral is equal to $$\iint_D \sqrt{x^2+y^2}\,dx\,dy$$

Where $D$ is a set (see here for plot) defined in polar coordinates as:

$$D:0<r<2,\quad \tfrac12\pi< \theta<\tfrac32\pi$$

Use the fact that $x^2+y^2=r^2$ and rewrite your integral as $$\int_{\tfrac12\pi}^{\tfrac32\pi}\int_0^2\sqrt{r^2}r\,dr\,d\theta$$

Answer 2

Once we identify the region of integration, things will be straightforward.

Note that $y=\sqrt{4-x^2}$ is the equation of the top half of the circle $x^2+y^2=4$, and $y=-\sqrt{4-x^2}$ is the equation of the bottom half. Draw the circle.

In the given double integral, after $y$ goes from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$, the variable $x$ goes from $-2$ to $0$. So we are integrating over the left half of the disk with centre the origin and radius $2$.

Now the bounds are can be read off. The variable $r$ goes from $0$ to $2$, and the angle $\theta$ goes from $\pi/2$ to $3\pi/2$.

Answer 3

To transform coordinates from $(x,y)$ to $(r,\theta)$, we must calculate the Jacobian. This is the determinant

$$\begin{Vmatrix} \frac{\partial x}{\partial r}&\frac{\partial y}{\partial r}\\\frac{\partial x}{\partial \theta}&\frac{\partial y}{\partial \theta}\end{Vmatrix}=\begin{Vmatrix} \cos \theta &\sin \theta\\-r\sin \theta & r\cos \theta \end{Vmatrix}=r$$

Therefore, $dy\,dx\to r\,dr\,d\theta$.

Next, we calculate the integrand. Here, we have $$\sqrt{x^2+y^2}=\sqrt{(r\cos \theta)^2+(r\sin \theta)^2}=r$$

Finally, we examine the region of integration in the $x-y$ plane. We see that it is the region in the left-half plane, of the disk centered at the origin and with radius $2$. In the $r-\theta$ plane, this corresponds to a rectangular region $0<r<2$ and $\pi/2< \theta <3\pi/2$.

Putting everything together reveals

$$\int_{-2}^{0}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\sqrt{x^2+y^2}\,dy\,dx=\int_{\pi/2}^{3\pi/2}\int_{0}^{2}(r)\, r\,dr\,d\theta=\frac83 \pi$$