I have some difficulties in solving this problem:
Find the area of a figure, specified by the inequalities: $x^2 + y^2 \leq 2x$ and $x^2 + 2x + y^2 \leq 3$
I know that I have to use the formula for area/surface of a curvilinear trapezoid but don't know exactly how?
Any ideas would be greatly appreciated.
p.s. Sorry for my bad English
Regarding to @lab's comment, you see that the area has a good symmetric. So it is enough to compute the colored area and then making the result doubled. See:
The brown area: $$\int_0^{3/4}\int_0^{\sqrt{2x-x^2}}dydx=\int_0^{3/4}\sqrt{2x-x^2}dx $$
The blue area: $$\int_{3/4}^2\int_0^{\sqrt{3-2x-x^2}}dydx=\int_{3/4}^2\sqrt{3-2x-x^2}dx$$