# 2003 AP BC Calculus (Form b) Free-Response Question #2

The graphs of the circles $x^2+y^2 = 2$ and $\left(x -1\right)^2+y^2=1$. The graphs intersect at the point (1,1) and (1,-1). Let $R$ be the region in the first quadrant bounded by the two circles and the $x$-axis. (b) Set up an expression involving one or more integrals with respect to $y$ that represents the area of $R$.

$\int_0^1 \sqrt{2-y^2} - \left(1 + \sqrt{1-y^2}\right)\; \mathrm{d}y$ is my answer for (b) but the solution guide states $\int_0^1 \sqrt{2-y^2} - \left(1 - \sqrt{1-y^2}\right)\; \mathrm{d}y$

Louis A. Talman agrees with me.

My question Is my answer incorrect? I don't understand how the AP Board obtained their answer. Why would they take the negative root of the square root?

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... what is the question? –  Neal Jan 7 '13 at 12:59
@Neal should be something like set up the integrals and calculate the area. –  007resu Jan 7 '13 at 13:02
@user1710036 I don't need to calculate the area, just wondering why my answer differs from the solution guide of the ap exam. –  yiyi Jan 7 '13 at 13:08
What does (b) ask? –  Arthur Fischer Jan 7 '13 at 13:12

The only thing you need is the area between the green curve and blue curve. So Your answer is incorrect. The graph below will explain everything. You have to be careful while taking square root of the expression after $(x-1)^2=1-y^2\implies x=1 \pm \sqrt{(1-y^2)}$ Here you took $x= 1 + \sqrt{(1-y^2)}$ which is not the half of the circle needed here. So you should take $x= 1 - \sqrt{(1-y^2)}$
The set closest to the origin is $$R_1=\{(x,y):0\leq y\leq1, 1-\sqrt{1-y^2}\leq x\leq \sqrt{2-y^2}\}$$ and the other is $$R_2=\{(x,y):0\leq y\leq1, \sqrt{2-y^2}\leq x\leq 1+\sqrt{1-y^2}\}$$
Thus the area of $R_1$ is $$A_1 =\int_0^1\int_{1-\sqrt{1-y^2}}^{\sqrt{2-y^2}}dxdy=\int_0^1 \sqrt{2-y^2} - (1-\sqrt{1-y^2}) dy$$ while the area of $R_2$ is $$A_2 =\int_0^1\int_{\sqrt{2-y^2}}^{1+\sqrt{1-y^2}}dxdy=\int_0^1 (1+\sqrt{1-y^2})-\sqrt{2-y^2} dy$$