Find the volume of solid formed by the revolution of the curve $y^2(a+x)=x^2(a-x)$ about $X-$ axis.
My Attempt:
Here the equation of curve is
$$y^2(a+x)=x^2(a-x)$$
It is symmetrical on $X-$ axis and passes through origin and $(a,0)$. The volume generated by revolution of curve about $X-$ axis is defined as:
$$V=\int_{0}^{a} \pi y^2 dx$$
$$V=\pi \int_{0}^{a} \frac {x^2(a-x)}{a+x} dx$$
How to evaluate this integral?
Best Answer
With your integral of
$$V=\pi \int_{0}^{a} \frac {x^2(a-x)}{a+x} dx \tag{1}\label{eq1A}$$
you can use a substitution of $u = a + x \implies du = dx$. Also, $x = u - a$ and $a - x = a - (u - a) = 2a - u$, gives
$$\begin{equation}\begin{aligned} V & = \pi \int_{a}^{2a} \frac{(u - a)^2(2a-u)}{u}du \\ & = \pi \int_{a}^{2a} \frac{(u^2 - 2au + a^2)(2a-u)}{u}du \\ & = \pi \int_{a}^{2a} \frac{-u^3 + 4au^2 - 5a^2u + 2a^3}{u}du \\ & = \pi\left(-\int_{a}^{2a}u^2du + 4a\int_{a}^{2a}udu - 5a^2\int_{a}^{2a}du + 2a^3\int_{a}^{2a}\frac{1}{u}du\right) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
I'll leave it to you to do the final integrations, substitutions and simplifications.