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.

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