# Find the area/surface of a figure, specified by inequalities

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

• We have $x^2+y^2\le 2x\iff (x-1)^2+y^2\le1$ and $x^2+2x+y^2\le 3\iff (x+1)^2+y^2\le 2^2$ – lab bhattacharjee Sep 1 '13 at 6:44
• wow, very fast hint, thanks. But what I have to do next? – NoSense Sep 1 '13 at 7:11

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: $$\int_{3/4}^2\int_0^{\sqrt{3-2x-x^2}}dydx=\int_{3/4}^2\sqrt{3-2x-x^2}dx$$

• sorry but I don't know how to solve double integrals :( . Is there another way? – NoSense Sep 1 '13 at 7:08
• @NoSense: See the edit. – mrs Sep 1 '13 at 7:24
• sorry for the stupid questions, but how have you get to these integrals and how have you defined the two boundaries? – NoSense Sep 1 '13 at 7:32
• Could you help me, please? I really need to understand how to solve similar problems. – NoSense Sep 1 '13 at 8:34
• @NoSense: Dear friend, If I want to tell you what is going on here completely, I should teach you all the prerequisites before entering to Calculus II. :-) But, a simple point. Wherever we want to evaluate the region $a\leq x\leq b$ below the graph of continuous function $f$, then we evaluate $\int_a^b f dx$ – mrs Sep 1 '13 at 11:09