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Find the area of the circle x^2+y^2=16 by the method of intregration

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Answer:

Hello,


16\pi

Explanation:


I=(Area)/(4) =\int\limits^4_0 {√(16-x^2) } \, dx \\\\Let\ say\ x=4*sin(t),\ dx=4*cos(t) dt\\\\\displaystyle I=\int\limits^(\pi )/(2) _0 {4*√(1-sin^2(t)) }*4*cos(t) \, dt \\\\=16*\int\limits^(\pi )/(2) _0 {cos^2(t)} \, dt \\\\=16*\int\limits^(\pi )/(2) _0 {(1-cos(2t))/(2)} \, dt \\\\=8*[t]^(\pi )/(2) _0-[(sin(2t))/(2) ]^(\pi )/(2) _0\\\\=4\pi -0\\\\=4\pi\\\\\boxed{Area=4*I=16\pi}\\

User Luke Baughan
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