Convert the double integral into polar

Question

$$\int^{4}_{0}\int_{0}^{\sqrt{4x-x^2}}\sqrt{x^2+y^2}\text{ dy dx}=14.22$$

In polar, we have $0\leq y\leq \sqrt{4x-x^2}$

Simplying, we get $y^2=4x-x^2$, which turns into $(x-2)^2+y^2=4$. Here is the region we are considering. Clearly $0\leq\theta\leq\pi$, and we know that $x^2+y^2=4x$, which means $r^2=4r\cos\theta\to r=4\cos\theta$

Therefore, $0\leq r\leq 4\cos\theta$, and the original function $\sqrt{x^2+y^2}$ simply becomes $r$. So our integral now is:

$$\int^{\pi}_{0}\int^{4\cos\theta}_{0}r\text{ r dr d$\theta$}=\int^{\pi}_{0}\int^{4\cos\theta}_{0}r^2\text{ dr d$\theta$}=\int^{\pi}_{0}\left[\dfrac{r^3}{3}\right]^{4\cos\theta}_{0}\text{ d$\theta$}$$

$$=\dfrac{64}{3}\int^{\pi}_{0}\cos^3\theta\text{ d$\theta$}=0$$ But according to WA^, the cartesian coordinate integral is different?

Answer 1

Your range of angle is wrong.

The region doesn't appear in the second quadrant at all.

The relevant angle is $0 \leq \theta \leq \frac{\pi}{2}$.

Answer 2

You have to integrate $d\theta$ from $0$ to $\pi/2$.

This will give you

$$\int_0^{\pi/2}\cos^3\theta\ d\theta = \frac{2}{3}$$

Hence

$$\frac{128}{9} = 14.22(...)$$