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Q.
The area of the region bounded by the curve $y=\sqrt{16-x^2}$ and $X$-axis is
Application of Integrals
Solution:
We have, $y=\sqrt{16-x^2}$
On squaring both sides, we get
$y^2 =16-x^2 $
$x^2+y^2 =16$
Let us sketch the figure of the curve which represents a circle. $(y \geq 0)$
$\therefore$ Area of shaded region
$ =$ Area $(B O A C B)$
$=\int\limits_{-4}^4 \sqrt{16-x^2} d x$
$ =\int\limits_{-4}^4 \sqrt{(4)^2-x^2} d x $
$ =\left[\frac{x}{2} \sqrt{16-x^2}+8 \sin ^{-1}\left(\frac{x}{1}\right)\right]_{-4}^4$
$=\left[\left\{\frac{4}{2} \sqrt{16-16}+8 \sin ^{-1}\left(\frac{4}{4}\right)\right\}\right.$
$\left.-\left\{\frac{-4}{2} \sqrt{16-(-4)^2}+8 \sin ^{-1}\left(\frac{-4}{4}\right)\right\}\right]$
$=\left[2 \times 0+8 \times \frac{\pi}{2}+8 \times \frac{\pi}{2}\right]$
$=8 \pi$ sq units
