The area of the shaded region in the given fi | vedclass

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Question

(A) The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$,which can be written as $\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$. Here,$a=4$ and $b=3$.

Vedclass · Accepted Answer

$3(\pi-2)$

Vedclass · Answer

(A) The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$,which can be written as $\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$. Here,$a=4$ and $b=3$.From the figure,the shaded region is bounded by the ellipse in the first quadrant,specifically between $x=0$ and $x=4$,and the line connecting $(0,3)$ and $(4,0)$.The equation of the line passing through $(0,3)$ and $(4,0)$ is $\frac{x}{4}+\frac{y}{3}=1$,which gives $y = \frac{3}{4}(4-x)$.The area of the shaded region is the area under the ellipse minus the area under the line in the first quadrant.Area $= \int_{0}^{4} \frac{3}{4}\sqrt{16-x^2} \, dx - \int_{0}^{4} \frac{3}{4}(4-x) \, dx$.$= \frac{3}{4} \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}(\frac{x}{4}) \right]_{0}^{4} - \frac{3}{4} \left[ 4x - \frac{x^2}{2} \right]_{0}^{4}$.$= \frac{3}{4} [0 + 8(\frac{\pi}{2})] - \frac{3}{4} [16 - 8] = 3\pi - 6 = 3(\pi-2)$ sq. units.

The area of the shaded region in the given figure of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is . . . . . . sq. units.

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