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fazi178
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Tyro
Asked: 2021-03-29T12:18:16+00:00 In:

Area of a circle thrpugh integration

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what is the area of circle using integration method

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  2. acedstud
    This answer was edited.

    The equation of a circle with radius, r, is given by x^2+y^2=r^2.

    Making y the subject of the formula, we have y^2=r^2-x^2\Rightarrow y=\pm\sqrt{r^2-x^2}.

    Now, the radius of a circle divides the circle into four equal parts called the quadrants. And the area of the portion bounded by a curve between two points on the x-axis is given by the integral between the two points of the curve.

    Thus, the area of a quadrant of a circle bounded by the center of the circle and the circumference (that is, the radius) is given by A=\int_0^r{\sqrt{r^2-x^2}}dx        [Note that Area is always positive so we take the positive value of y.]

    Thus, the area of the 4 quadrants of a circle is given by A=4\int_0^r{\sqrt{r^2-x^2}}dx.

    Converting to polar coordinates, x=r\cos\theta\Rightarrow dx=-r\sin\theta d\theta. When x=0, \cos\theta=0\Rightarrow\theta=\cos^{-1}{0}=\dfrac{\pi}{2} and when x=r, \cos\theta=1\Rightarrow\theta=\cos^{-1}{1}=0

    Thus, A=4\int_{\frac{\pi}{2}}^0{\sqrt{r^2-r^2\cos^2\theta}}\cdot-r\sin\theta d\theta

    =-4\int_0^{\frac{\pi}{2}}{\sqrt{r^2(1-\cos^2\theta)}}\cdot-r\sin\theta d\theta

    =-4\int_0^{\frac{\pi}{2}}{r\sqrt{\sin^2\theta)}}\cdot-r\sin\theta d\theta        [\because using the Pythagorean trigonometric identity \sin^2\theta+\cos^2\theta=1\Rightarrow1-\cos^2\theta=\sin^2\theta.]

    =-4\int_0^{\frac{\pi}{2}}{-r^2\sin^2\theta}d\theta=4r^2\int_0^{\frac{\pi}{2}}{\sin^2\theta}d\theta

    Using the two trigonometric identities \sin^2\theta+\cos^2\theta=1 and \cos^2\theta-\sin^2\theta=\cos{2\theta}, we have that \sin^2\theta=\dfrac{1}{2}(1-\cos{2\theta}).

    Thus we have, A=4r^2\int_0^{\frac{\pi}{2}}{\dfrac{1}{2}(1-\cos{2\theta})}d\theta

    =2r^2\int_0^{\frac{\pi}{2}}{(1-\cos{2\theta})}d\theta

    =2r^2\left[\theta-\dfrac{\sin{2\theta}}{2}\right]_0^{\frac{\pi}{2}}

    =2r^2\left[\left(\dfrac{\pi}{2}-\dfrac{\sin{2\cdot\frac{\pi}{2}}}{2}\right)-\left(0-\dfrac{\sin{2(0)}}{2}\right)\right]

    =2r^2\left(\dfrac{\pi}{2}-\dfrac{\sin\pi}{2}+\dfrac{\sin{0}}{2}\right)

    =2r^2\left(\dfrac{\pi}{2}\right)        [\because \sin\pi=\sin0=0]

    =\pi r^2

    Therefore, the area of a circle with a radius of r is given by A=\pi r^2.

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