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

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(B) The given curve is $x = 4 - y^2$, which is a parabola opening to the left with its vertex at $(4, 0)$.To find the intersection points with the $Y$

Vedclass · Accepted Answer

$\frac{32}{3} 	ext{ sq. unit}$

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(B) The given curve is $x = 4 - y^2$, which is a parabola opening to the left with its vertex at $(4, 0)$.To find the intersection points with the $Y$-axis, we set $x = 0$:$0 = 4 - y^2 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.Thus, the curve intersects the $Y$-axis at $(0, 2)$ and $(0, -2)$.The area bounded by the curve and the $Y$-axis is symmetric about the $X$-axis.Area $= \int_{-2}^{2} x \, dy = \int_{-2}^{2} (4 - y^2) \, dy$.Due to symmetry, Area $= 2 \int_{0}^{2} (4 - y^2) \, dy$.$= 2 \left[ 4y - \frac{y^3}{3} ight]_{0}^{2}$.$= 2 \left( (4(2) - \frac{2^3}{3}) - (0 - 0) ight)$.$= 2 \left( 8 - \frac{8}{3} ight) = 2 \left( \frac{24 - 8}{3} ight) = 2 \left( \frac{16}{3} ight) = \frac{32}{3} 	ext{ sq. unit}$.

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

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