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

1.	Find area of the circle x^2+y^2=36 by using definite integration
Answer» $\underline{\textbf{Given:}}$ $\textsf{Circle is}\;\mathsf{x^<klux>2</klux>+y^2=36}$ $\underline{\textbf{To find:}}$ $\textsf{Area of the given circle}$ $\underline{\textbf{Solution:}}$ $\mathsf{Put\;y=<klux>0</klux>,}$ $\mathsf{x^2=36\implies\;x=\pm\,6}$ $\therefore\textsf{The circle meets x-axis at (-6,0) and (6,0)}$ $\textsf{Similarly, the circle meets the y-axis at (0,6) and (0,-6)}$ $\textsf{Since the circle is symmetric with respect to both the axes,}$ $\mathsf{Area\;o\;the\;circle}$ $\mathsf{=4{\times}Area\;bounded\;by\;the\;circle\;in\;I\;quadrant}$ $\therefore\mathsf{=4{\times}\displaystyle\int\limits^6_0\;y\;dx}$ $\therefore\mathsf{=4{\times}\displaystyle\int\limits^6_0\;\sqrt{36-x^2}\;dx}$ $\therefore\mathsf{=4{\times}\displaystyle\int\limits^6_0\;\sqrt{6^2-x^2}\;dx}$ $\textsf{Using the formula,}$ $\boxed{\mathsf{\int\;\sqrt{a^2-x^2}\;dx=\dfrac{x\sqrt{a^2-x^2}}{2}+\dfrac{a^2}{2}\,sin^{-1}\left(\dfrac{x}{a}\right)+C}}$ $\mathsf{=4\left[\dfrac{x\sqrt{6^2-x^2}}{2}+\dfrac{6^2}{2}\,sin^{-1}\left(\dfrac{x}{6}\right)\right]^6_0}$ $\mathsf{=4\left[\dfrac{x\sqrt{6^2-6^2}}{2}+\dfrac{6^2}{2}\,sin^{-1}\left(\dfrac{6}{6}\right)-(0+0)\right]}$ $\mathsf{=4\left[0+18\,\dfrac{\<klux>PI</klux>}{2}-(0)\right]}$ $\mathsf{=4\left[\dfrac{18\pi}{2}\right]}$ $\mathsf{=36\,\pi\;square\;units}$ $\underline{\textbf{Find more:}}$ LIMIT 0 to Pi /2 INTEGRATION xsinx brainly.in/question/5372600

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