The area bounded by the curve x=4-y^{2} and t | vedclass

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Question

(C) The given curve is $x = 4 - y^2$. The curve intersects the $Y$-axis where $x = 0$,which gives $4 - y^2 = 0$,so $y = \pm 2$. The points of intersec

Vedclass · Accepted Answer

$\frac{32}{3} 	ext{ sq units}$

Vedclass · Answer

(C) The given curve is $x = 4 - y^2$. The curve intersects the $Y$-axis where $x = 0$,which gives $4 - y^2 = 0$,so $y = \pm 2$. The points of intersection are $(0, 2)$ and $(0, -2)$.Since the curve is symmetric about the $X$-axis,the total area $A$ is twice the area in the first quadrant.$A = 2 \int_{0}^{2} x \, dy$$A = 2 \int_{0}^{2} (4 - y^2) \, dy$$A = 2 \left[ 4y - \frac{y^3}{3} ight]_{0}^{2}$$A = 2 \left[ (4(2) - \frac{2^3}{3}) - (0) ight]$$A = 2 \left[ 8 - \frac{8}{3} ight]$$A = 2 \left[ \frac{24 - 8}{3} ight] = 2 \left( \frac{16}{3} ight) = \frac{32}{3} 	ext{ sq units}$.

The area bounded by the curve $x=4-y^{2}$ and the $Y$-axis is

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