The area (in sq.

Question

The area (in sq. units) of the smaller portion enclosed between the curves $x^2+y^2=4$ and $y^2=3x$ is:

Accepted Answer

Solving $x^2+3x-4=0$ (from substituting $y^2=3x$ into the circle) gives $x=1$ (taking $x\ge0$), $y=\pm\sqrt3$.

$$\text{Area} = 2\left[\int_0^1\sqrt3\sqrt x\,dx + \int_1^2\sqrt{4-x^2}\,dx\right]$$

$$= 2\left[\sqrt3\cdot\frac23 + \left(\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\frac x2\right)_1^2\right] = 2\left[\frac{2\sqrt3}{3} + \pi - \frac{\sqrt3}{2}-\frac\pi3\right]$$

$$= \frac{1}{\sqrt3}+\frac{4\pi}{3}$$

Solutions · 1