The area between x=y^{2} and x=4 is divided i | vedclass

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Question

(A) The area bounded by the parabola $x=y^{2}$ and the line $x=4$ is symmetric about the $x$-axis.The line $x=a$ divides this area into two equal part

Vedclass · Accepted Answer

$4^{\frac{2}{3}}$

Vedclass · Answer

(A) The area bounded by the parabola $x=y^{2}$ and the line $x=4$ is symmetric about the $x$-axis.The line $x=a$ divides this area into two equal parts.Total area from $x=0$ to $x=4$ is $2 \int_{0}^{4} \sqrt{x} \, dx = 2 \left[ \frac{2}{3} x^{\frac{3}{2}} ight]_{0}^{4} = \frac{4}{3} (8) = \frac{32}{3}$.The area from $x=0$ to $x=a$ is half of the total area:$2 \int_{0}^{a} \sqrt{x} \, dx = \frac{1}{2} 	imes \frac{32}{3} = \frac{16}{3}$.$2 \left[ \frac{2}{3} x^{\frac{3}{2}} ight]_{0}^{a} = \frac{16}{3}$.$\frac{4}{3} a^{\frac{3}{2}} = \frac{16}{3}$.$a^{\frac{3}{2}} = 4$.$a = 4^{\frac{2}{3}}$.

The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$. Find the value of $a$.

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