Evaluate the double integral $\iint_D\sqrt{4-x^2-y^2}$ bounded by semi-circle

Question

I would appreciate it if someone can help me solve this question, as I'm struggling to get its answer.

Q: Evaluate the double integral $$\iint_D\sqrt{4-x^2-y^2}dxdy$$ bounded by semi-circle $$x^2+y^2=2x$$ and lying in first quadrant

Thanks

Answer 1

Using polar coordinates, $x^2+y^2=2x$ gives $r^2=2r\cos\theta$ so $r=2\cos\theta$.

For the part of the region inside the circle which is also in the first quadrant, we get

$\displaystyle\int_0^{\frac{\pi}{2}}\int_0^{2\cos\theta}\sqrt{4-r^2}\; r\;drd\theta=\int_0^{\frac{\pi}{2}}\left[-\frac{1}{3}(4-r^2)^{\frac{3}{2}}\right]_{r=0}^{r=2\cos\theta}d\theta=\int_0^{\frac{\pi}{2}}\frac{8}{3}(1-\sin^3\theta)d\theta$

$\displaystyle=\frac{8}{3}\left[\theta+\cos\theta-\frac{1}{3}\cos^3\theta\right]_0^{\frac{\pi}{2}}=\frac{8}{3}\left(\frac{\pi}{2}-\frac{2}{3}\right)=\frac{4}{9}\left(3\pi-4\right)$

Answer 2

We have: $$x^2+y^2=2x$$

Moving everything to the left: $$x^2-2x+y^2=0$$

Completing the square: $$(x-1)^2+y^2=1$$

Which is a circle with centre $(1,0)$, hence: $$0\le x\le2$$

Solving $y$: $$y=\sqrt{1-(x-1)^2}$$

So the integral becomes: $$\int_{x=0}^{x=2}\int_{y=0}^{y=\sqrt{1-(x-1)^2}}\sqrt{4-x^2-y^2}\ \mathrm dy\ \mathrm dx$$