The area in square units of the region bounde | vedclass

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(C) The given curve is $y = \sqrt{16 - x^2}$, which represents the upper semi-circle of $x^2 + y^2 = 4^2$ with radius $r = 4$.We need to find the area

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The given curve is $y = \sqrt{16 - x^2}$, which represents the upper semi-circle of $x^2 + y^2 = 4^2$ with radius $r = 4$.We need to find the area bounded by the curve, $x = 0$, $x = 4$, and the $X$-axis.This represents one-quarter of the area of the circle with radius $r = 4$.Area $= \int_{0}^{4} \sqrt{16 - x^2} \, dx$.Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C$.Area $= [\frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}(\frac{x}{4})]_{0}^{4}$.Area $= [\frac{4}{2} \sqrt{16 - 16} + 8 \sin^{-1}(1)] - [0 + 8 \sin^{-1}(0)]$.Area $= [0 + 8(\frac{\pi}{2})] - [0] = 4\pi$ square units.

The area in square units of the region bounded by the curve $y = \sqrt{16 - x^2}$ and the lines $x = 0, x = 4$ above the $X$-axis is: (in $\pi$)

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