The plane figure bounded by the parabola $y^2=4ax$, the x-axis and the ordinate at $x=a$, is rotated a complete revolution about the line $x=-a$. Find the volume of the solid generated.
I tried to integrate this by putting a and - a as the upper and lower limits but to no avail $$\pi\int_{a}^{a} 4ax\mathrm{dx}$$
How do I go about answering this question?
Then, what does it mean to rotate a curve about a line?
Express $y$ as a function of $x$ for the first quadrant: $y=2\sqrt{ax}$. Use the restriction that $a>0$ (for negative values of $a$, you would get the same volume because the solid is symmetric with respect to the axis of revolution) and shift the curve along the $x$-axis $a$ units to the right: $y=2\sqrt{a(x-a)}$. And then you're going to use shell integration with the bounds of $a$ and $2a$:
$$ V=2\pi\int_{a}^{2a}x\cdot 2\sqrt{a(x-a)}\,dx=4\pi\int_{a}^{2a}x\sqrt{ax-a^2}\,dx. $$
You can use the substitution $u=ax-a^2$ to solve that integral:
$$ \frac{4\pi}{a^2}\int_{a}^{2a}(ax-a^2+a^2)\sqrt{ax-a^2}\frac{d}{dx}(ax-a^2)\,dx=\\ \frac{4\pi}{a^2}\int_{0}^{a^2}(u+a^2)\sqrt{u}\,du= \frac{4\pi}{a^2}\int_{0}^{a^2}(u^{\frac{3}{2}}+a^2u^{\frac{1}{2}})\,du. $$
From this point on, the integration is trivial.
