Question

Find area of the circle x^2+y^2=36 by using definite integration

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Circle is}\;\mathsf{x^2+y^2=36}$

$\underline{\textbf{To find:}}$

$\textsf{Area of the given circle}$

$\underline{\textbf{Solution:}}$

$\mathsf{Put\;y=0,}$

$\mathsf{x^2=36\implies\;x=\pm\,6}$

$\therefore\textsf{The circle meets x-axis at (-6,0) and (6,0)}$

$\textsf{Similarly, the circle meets the y-axis at (0,6) and (0,-6)}$

$\textsf{Since the circle is symmetric with respect to both the axes,}$

$\mathsf{Area\;o\;the\;circle}$

$\mathsf{=4{\times}Area\;bounded\;by\;the\;circle\;in\;I\;quadrant}$

$\therefore\mathsf{=4{\times}\displaystyle\int\limits^6_0\;y\;dx}$

$\therefore\mathsf{=4{\times}\displaystyle\int\limits^6_0\;\sqrt{36-x^2}\;dx}$

$\therefore\mathsf{=4{\times}\displaystyle\int\limits^6_0\;\sqrt{6^2-x^2}\;dx}$

$\textsf{Using the formula,}$

$\boxed{\mathsf{\int\;\sqrt{a^2-x^2}\;dx=\dfrac{x\sqrt{a^2-x^2}}{2}+\dfrac{a^2}{2}\,sin^{-1}\left(\dfrac{x}{a}\right)+C}}$

$\mathsf{=4\left[\dfrac{x\sqrt{6^2-x^2}}{2}+\dfrac{6^2}{2}\,sin^{-1}\left(\dfrac{x}{6}\right)\right]^6_0}$

$\mathsf{=4\left[\dfrac{x\sqrt{6^2-6^2}}{2}+\dfrac{6^2}{2}\,sin^{-1}\left(\dfrac{6}{6}\right)-(0+0)\right]}$

$\mathsf{=4\left[0+18\,\dfrac{\pi}{2}-(0)\right]}$

$\mathsf{=4\left[\dfrac{18\pi}{2}\right]}$

$\mathsf{=36\,\pi\;square\;units}$

$\underline{\textbf{Find more:}}$

Limit 0 to Pi /2 integration xsinx

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