Browse Questions

Find the area of a minor segment of the circle $x^2+y^2=a^2$ cut off by the line $x=\large \frac{a}{2}.$

Toolbox:
• If we are given two curves represented by y=f(x);y=g(x),where $f(x)\geq g(x)$ in [a,b],the points of intersection of two curves are given by x=a and x=b,by taking common values of y from the equation of the two curves.
Step 1:
Let the equation of the curve $x^2+y^2=a^2$ be $R_1$ and the equation of the line $x=\large\frac{a}{2}$$=R_2 Let us find the point of intersection of the two curves by solving the two equation. x^2+y^2=a^2 x=\large\frac{a}{2} \Rightarrow \large\frac{a^2}{4}$$+y^2=a^2$
$\Rightarrow y^2=a^2-\large\frac{a^2}{4}=\frac{3a^2}{4}$
$\Rightarrow y=\pm\large\frac{\sqrt 3 a}{2}$ and $x=\pm \large\frac{a}{2}$
$\therefore P(\large\frac{x}{2},\frac{\sqrt 3a}{2})$ is the point of intersection in the first quadrant.
Step 2:
The smaller region bounded by the curves is the shaded portion as shown in the figure.
$\qquad=2\int\limits_{\sqrt 3a/2}^a\sqrt{a^2-x^2}dx$
On integrating we get,
$2\big[\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{1}{2}$$a^2\sin^{-1}\large\frac{x}{a}\big]_{\sqrt 3a/2}^a$
On applying limits we get,
$2[0+\large\frac{a^2}{2}$$\sin^{-1}(1)-[\large\frac{\sqrt 3a}{2}$$\sqrt{a^2-\large\frac{3a^2}{4}}+\large\frac{a^2}{2}$$\sin^{-1}(\large\frac{\sqrt 3a}{2a})] But \sin^{-1}(1)=\large\frac{\pi}{2} and \sin^{-1}(\large\frac{\sqrt 3}{2})=\large\frac{\pi}{3} \Rightarrow 2[\large\frac{a^2}{2}\frac{\pi}{2}-\frac{\sqrt 3a}{2}\times \frac{\sqrt 3a}{2}-\frac{a^2}{2}.\frac{\pi}{3}] \Rightarrow \large\frac{\pi a^2}{2}-\frac{3a^2}{2}-\frac{\pi a^2}{3} \Rightarrow \large\frac{\pi a^2}{6}-\frac{3a^2}{2} \Rightarrow \large\frac{a^2}{2}\big(\large\frac{\pi}{3}$$-3)$ sq.units