Question

What is the area of the region bounded by the parabola $y^2=6(x-1)$ and $y^2=3x$

• A. $\frac{\sqrt{6}}{3}$
• B. $\frac{2\sqrt{6}}{3}$
• C. $\frac{4\sqrt{6}}{3}$
• D. $\frac{5\sqrt{6}}{3}$

Answer 1

The correct option is C $\frac{4\sqrt{6}}{3}$

Solving $y^2=6(x-1)$ and $y^2=3x$,

We get $6x-6=3⟹x=2⟹y=\pm \sqrt{6}$

$\therefore$ Area=${\int }_{-\sqrt{6}}^{\sqrt{6}}(1+\frac{y^2}{6}-\frac{y^2}{3})dy=2{\int }_0^{\sqrt{6}}(1-\frac{y^2}{6})dy=\frac{4\sqrt{6}}{3}$