# Problems

The figure below shows the graphs of the circles ${{x}^{2}}+{{y}^{2}}=2$ and ${{(x-1)}^{2}}+{{y}^{2}}=1$ The graphs intersect at the point $(1,1)$ and $(1,-1)$. Let $R$ be the shaded region in the first quadrant bounded by the two circles and the $x$-axis.

(a) Set up, but do not evaluate, an expression involving one or more integral with respect to x that represent the area of $R$.

(b) Set up, but do not evaluate, an expression involving one or more integral with respect to y that represent the area of $R$.

(c) The polar equation of the circles are ﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿$r = \sqrt 2$ and $r=\cos \theta$, respectively. Set up, but do not evaluate, an expression involving one or more integrals with respect to the polar $\theta$ that represent area of $R$.

For part (a), we express both circles as functions of $x$, giving $y = f_1(x) = \sqrt{2 - x^2}$, $\quad y = f_2(x) = \sqrt{1 - (x-1)^2}$, where we take the positive square root since we are only interested in $y \ge 0$.  Then clearly, over $x \in [0,1)$, we have $f_1(x) > f_2(x)$, and over $(1, \sqrt{2}]$, we have $f_1(x) < f_2(x)$.  Thus the area of $R$ is given piecewise as $R= \int_{x=0}^1 f_2(x)dx + \int_{x=1}^{\sqrt{2}} f_1(x)\,$, $= \int_{x=0}^1 \sqrt{1-(x-1)^2} \, dx + \int_{x=1}^{\sqrt{2}} \sqrt{2-x^2} \, dx.$

For part (b), we express both circles as functions of $y$, giving $x = g_1(y) = \sqrt{2 - y^2}$, $\quad x = g_2(y) = 1 - \sqrt{1-y^2}$,  where we have carefully chosen the negative root for the second function because this represents the part of the circle that lies to the left of $x = 1$.  Thus for $y \in [0,1)$, we have $g_1(y) > g_2(y)$, and the area is simply $R= \int_{y=0}^1 g_1(y) - g_2(y) dy = \int_{y=0}^1 \sqrt{2 - y^2} - (1 - \sqrt{1-y^2}) dy$

For part (c), we use the normal area formula of polar curve:

So we get $R=\frac{1}{2}\int_{\frac{\pi }{4}}^{\frac{7\pi }{4}}{{{(x\cos \theta )}^{2}}d\theta }+2.\frac{1}{2}\int_{0}^{\frac{\pi }{4}}{{{(\sqrt{2})}^{2}}d\theta }$

As for the actual value of $R$, one can use either integral described above to find $R = (\pi - 1)/2$, but a simpler method utilizes basic geometry.  We first observe that the radius joining the point of intersection $A = (1,1)$ to the origin forms a 45-degree angle with the positive $x$-axis.  Also, the radius joining the same point of intersection to the center of the smaller circle at $B = (1,0)$ forms a 90-degree angle with the $x$-axis.  This suggests that we draw the right $\triangle OAB$, and consider the three regions into which this triangle partitions $R$.  Then the area of $R$ is immediately seen to be the sum of an eighth of the larger circle’s area plus a quarter of the smaller’s, minus the area of $\triangle OAB$:  $R= \frac{1}{8} \pi (\sqrt{2})^2 + \frac{1}{4} \pi - \frac{1}{2} = \frac{\pi - 1}{2},$ as claimed.