Question

Find the area of the circle x² + y² =a² ( by integration.
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Answer 1

We're given the circle,

$x^2+y^2=a^2$

First we have to rewrite it by taking y as a function in x.

$y^2=a^2-x^2\\\\\\y=\sqrt {a^2-x^2}\\\\\\y=a\sqrt {1-\left (\dfrac {x^2}{a^2}\right)}$

Actually this circle is centered at the origin (0, 0) having radius a. This circle is, so, symmetric along x and y axes. So, we may take four times the area of one - fourth the circle by integration to get the area of circle. So,

$\displaystyle \int y\ dx=a\int\limits_0^a\sqrt {1-\left(\dfrac {x^2}{a^2}\right)}\ dx$

Let,

$\dfrac {x}{a}=\sin\theta\\\\\\x=a\sin\theta\\\\\\dx=a\cos\theta\ d\theta$

Then,

$\displaystyle\int y\ dx=a^2\int\limits_0^{\frac {\pi}{2}}\cos\theta\sqrt {1-\sin^2\theta}\ d\theta\\\\\\\int y\ dx=a^2\int\limits_0^{\frac {\pi}{2}}\cos^2\theta\ d\theta\\\\\\\int y\ dx=a^2\int\limits_0^{\frac {\pi}{2}}\dfrac {1+\cos (2\theta)}{2}\ d\theta\\\\\\\int y\ dx=\dfrac {a^2}{2}\int\limits_0^{\frac {\pi}{2}}1\ d\theta+\dfrac {a^2}{2}\int\limits_0^{\frac {\pi}{2}}\cos (2\theta)\ d\theta$

$\displaystyle\int y\ dx=\dfrac {a^2}{2}\left [\theta\right]_0^{\frac {\pi}{2}}+\dfrac {a^2}{4}\int\limits_0^{\frac {\pi}{2}}\cos (2\theta)\ d(2\theta)\\\\\\\int y\ dx=\dfrac {a^2}{2}\left (\dfrac {\pi}{2}-0\right)+\dfrac {a^2}{4}\left [\sin (2\theta)\right]_0^{\frac {\pi}{2}}\\\\\\\int y\ dx=\dfrac {a^2}{2}\cdot\dfrac {\pi}{2}+\dfrac {a^2}{4}\left (\sin\pi-\sin 0\right)\\\\\\\int y\ dx=\dfrac {\pi a^2}{4}$

This is one fourth of the area of the circle. Hence total area is,

$4\cdot\dfrac {\pi a^2}{4}=\pi a^2$