The area of the region bounded by the curve y = 16 - x 2 and X -axis is:

Question

The area of the region bounded by the curve $y = \sqrt{16 - x^{2}}$ and $X$ -axis is:

Smart Achievers · Accepted Answer

(1)

Ans. $8 π$ sq. units

Explanation:

Given equation of curve is $y = \sqrt{16 - x^{2}}$ and the equation of line is $X$ -axis i.e., $y = 0 .$

$∴$ $\sqrt{16 - x^{2}} = 0$ ...(i)

$\Rightarrow$ $16 - x^{2} = 0$

$\Rightarrow$ $x^{2} = 16 \Rightarrow x = \pm 4$

So, the intersection points are (4, 0) and $(- 4, 0)$ .

Therefore, area of the curve,

$A = \int_{- 4}^{4} \sqrt{16 - x^{2}} d x$

$= \int_{- 4}^{4} \sqrt{4^{2} - x^{2}} d x$

$= {[\frac{x}{2} \sqrt{4^{2} - x^{2}} + \frac{4^{2}}{2} \sin^{- 1} (\frac{x}{4})]}_{- 4}^{4}$

$= [\frac{4}{2} \sqrt{4^{2} - 4^{2}} + 8 \sin^{- 1} (\frac{4}{4})] - [- \frac{4}{2} \sqrt{4^{2} - {(- 4)}^{2}} + 8 \sin^{- 1} (- \frac{4}{4})]$

$= [2 \cdot 0 + 8 \cdot \frac{π}{2} - 0 + 8 \cdot \frac{π}{2}] = 8 π sq. units.$

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