The area bounded by the curve 4y^2 = x^2(4-x)(x-2) is equal to:

Question

The area bounded by the curve $4y^2 = x^2(4-x)(x-2)$ is equal to:

Accepted Answer

$4y^2=x^2(4-x)(x-2)$ requires $(4-x)(x-2)\ge0$, i.e. $x\in[2,4]$. Write $y = \pm\dfrac{x}{2}\sqrt{(4-x)(x-2)}$ for $x>0$.

$$\text{Area} = \int_2^4 x\sqrt{(4-x)(x-2)}\,dx$$

Using the property $\displaystyle\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ with $a+b=6$:

$$\text{Area} = \int_2^4(6-x)\sqrt{(4-x)(x-2)}\,dx$$

Adding the two forms: $2\cdot\text{Area} = 6\displaystyle\int_2^4\sqrt{(4-x)(x-2)}\,dx = 6\int_2^4\sqrt{1-(x-3)^2}\,dx$

$$\text{Area} = 3\left[\frac{x-3}{2}\sqrt{1-(x-3)^2}+\frac12\sin^{-1}(x-3)\right]_2^4 = 3\left(\frac{\pi}{4}+\frac{\pi}{4}\right) = \frac{3\pi}{2}$$

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