The area bounded by the circle x^{2}+y^{2}=16 | vedclass

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(A) The given equation of the circle is $x^{2}+y^{2}=16$.Since the circle is symmetric about both the $x$-axis and $y$-axis,the total area bounded by

Vedclass · Accepted Answer

$\left[4 \sqrt{3}+\frac{8 \pi}{3}\right]$ sq. units

Vedclass · Answer

(A) The given equation of the circle is $x^{2}+y^{2}=16$.Since the circle is symmetric about both the $x$-axis and $y$-axis,the total area bounded by the lines $x=0$ and $x=2$ is twice the area of the region in the first quadrant.$	ext{Area} = 2 \int_{0}^{2} y \, dx = 2 \int_{0}^{2} \sqrt{16-x^{2}} \, dx$.Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}ight)$:$	ext{Area} = 2 \left[ \frac{x}{2} \sqrt{16-x^{2}} + \frac{16}{2} \sin^{-1} \left(\frac{x}{4}ight) ight]_{0}^{2}$$= 2 \left[ \left( \frac{2}{2} \sqrt{16-4} + 8 \sin^{-1} \left(\frac{2}{4}ight) ight) - (0 + 8 \sin^{-1}(0)) ight]$$= 2 \left[ \sqrt{12} + 8 \sin^{-1} \left(\frac{1}{2}ight) ight]$$= 2 \left[ 2\sqrt{3} + 8 \left(\frac{\pi}{6}ight) ight]$$= 4\sqrt{3} + 8 \left(\frac{\pi}{3}ight) = 4\sqrt{3} + \frac{8\pi}{3} 	ext{ sq. units.}$

The area bounded by the circle $x^{2}+y^{2}=16$ and the lines $x=0$ and $x=2$ is

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