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January 4th, 2018, 11:10 AM   #1
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Find area

Evaluate area of the region enclosed by the circle x^2+y^2=4 that which is above the line y=1 and below the line y=sqrt(3)x
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January 4th, 2018, 11:54 AM   #2
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Have you drawn a graph? $x^2+ y^2= 4$ is, as you say, a circle, with center at (0, 0) and radius 2. $y= \sqrt{3}x$ is a straight line. That line intersects y= 1 where $\sqrt{3}x= 1$ so $x= \frac{1}{\sqrt{3}}= \frac{\sqrt{3}}{3}$. It intersects the circle where $x^2+ 3x^2= 4x^2= 4$ so x= 1.

The region inside that circle, above y= 1 and under $y= \sqrt{3}x$ can be broken into two parts. In the first, for $0\le x\le \frac{\sqrt{3}}{3}$, y ranges from 1 to $\sqrt{3}x$. The area of that region is $\sqrt{3}\int_0^{\frac{\sqrt{3}}{3}} x- 1 dx$ which is, of course, the area of the triangle with base $\frac{\sqrt{3}}{3}$ and height $\sqrt{3}- 1$. The area of the second region, from $x= \frac{\sqrt{3}}{3}$ to $x= \sqrt{3}$, y from 1 to $y= \sqrt{4- x^2}$, is $\int_{\frac{\sqrt{3}}{3}}^\sqrt{3} \sqrt{4- x^2}dx$.

The entire area is the sum or those two.
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January 4th, 2018, 01:26 PM   #3
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$\displaystyle A = \int_{\pi/6}^{\pi/3} 2 - \dfrac{\csc^2{\theta}}{2} \, d\theta$
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