Answer to Question #218806 in Calculus for Dhwanitw

Question

Find the area of the curve y2 (2a – x) = x3
between the area and its asymptotes.

Answer 1

The asymptote(s) of the curve parallel to $y$ -axis is given by $2a-x=0.$

Then the vertical asymptote is $x=2a.$

$$y=\sqrt{\dfrac{x^3}{2a-x}}$$

Let $x=2a\sin^2\theta.$

$$dx=4a\sin \theta \cos \theta d\theta$$

$$Area=A=\displaystyle\int_{0}^{2a}\sqrt{\dfrac{x^3}{2a-x}}dx$$

$$=\displaystyle\int_{0}^{\pi/2}\sqrt{\dfrac{8a^3\sin^6\theta}{2a-2a\sin^2\theta}}4a\sin \theta \cos \theta d\theta$$

$$=8a^2\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^3\theta\sin \theta \cos \theta}{\cos \theta}d\theta$$

$$=8a^2\displaystyle\int_{0}^{\pi/2}\sin^4\theta d\theta=2a^2\displaystyle\int_{0}^{\pi/2}(1-\cos(2\theta))^2 d\theta$$

$$=2a^2\displaystyle\int_{0}^{\pi/2}(1-2\cos(2\theta)+\cos^2(2\theta)) d\theta$$

$$=a^2\displaystyle\int_{0}^{\pi/2}(2-4\cos(2\theta)+1+\cos(4\theta)) d\theta$$

$$=a^2\bigg[3\theta-2\sin(2\theta)+\dfrac{1}{4}\sin4\theta()\bigg]\begin{matrix} \pi/2 \\ 0 \end{matrix}$$

$$=\dfrac{3\pi a^2}{2} (units^2)$$

$Area=\dfrac{3\pi ^2}{2}$ square units.