Question

If the area bounded by the curve $x=y^{2}$ and the line $x=4$ is divided into two equal parts by the line $x=a$, find the value of $a$.

Hint:

For regions symmetric about the $x$-axis, double the integral from $0$ to the positive bound.

Answer 1

Step 1: Total bounded area. The region is bounded between $x=y^{2}$ and $x=4$, symmetric about the $x$-axis. So, \[ \text{Area} = 2 \int_{0}^{2} \big(4 - y^{2}\big)\, dy \] \[ = 2\left[4y - \frac{y^{3}}{3}\right]_{0}^{2} = 2\left(8 - \frac{8}{3}\right) = 2 \cdot \frac{16}{3} = \frac{32}{3}. \]

Step 2: Half of the area. Each part should have area \[ \frac{1}{2}\cdot \frac{32}{3} = \frac{16}{3}. \]

Step 3: Area between $x=y^{2$ and $x=a$.} \[ \text{Required area} = 2 \int_{0}^{\sqrt{a}} (a - y^{2})\, dy \] \[ = 2\left[ay - \frac{y^{3}}{3}\right]_{0}^{\sqrt{a}} = 2\left(a\sqrt{a} - \frac{a^{3/2}}{3}\right) = 2\left(\frac{2a^{3/2}}{3}\right) = \frac{4}{3}a^{3/2}. \]

Step 4: Equating to half area. \[ \frac{4}{3}a^{3/2} = \frac{16}{3} \] \[ a^{3/2} = 4 \;\; $\Rightarrow$ \;\; a = 4^{2/3}. \]

Final Answer: \[ \boxed{a = 4^{2/3}} \]