Area of the region bounded by the curve x^2 = | vedclass

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Question

(A) The curve is $x^2 = 4y$, which implies $x = \pm 2\sqrt{y}$.
The region is bounded by the parabola and the line $y = 3$.
The area $A$ is given by

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The curve is $x^2 = 4y$, which implies $x = \pm 2\sqrt{y}$.
The region is bounded by the parabola and the line $y = 3$.
The area $A$ is given by the integral of the width of the parabola with respect to $y$ from $y = 0$ to $y = 3$.
$A = \int_{0}^{3} (x_{	ext{right}} - x_{	ext{left}}) \, dy = \int_{0}^{3} (2\sqrt{y} - (-2\sqrt{y})) \, dy = \int_{0}^{3} 4\sqrt{y} \, dy$.
$A = 4 \int_{0}^{3} y^{1/2} \, dy = 4 \left[ \frac{y^{3/2}}{3/2} ight]_{0}^{3} = 4 \cdot \frac{2}{3} [y^{3/2}]_{0}^{3}$.
$A = \frac{8}{3} (3^{3/2}) = \frac{8}{3} (3\sqrt{3}) = 8\sqrt{3}$.
Since $8\sqrt{3}$ is not explicitly in the options provided in the prompt but $4\sqrt{3}$ is often a distractor for half-area, we re-verify the calculation. The total area is $8\sqrt{3}$. Given the options, if the question intended the area in the first quadrant only, it would be $4\sqrt{3}$.

Area of the region bounded by the curve $x^2 = 4y$ and the line $y = 3$ is . . . . . . . (in $\sqrt{3}$)

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