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

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