From Demidovich:
A right parabolic segment whose base is $2a$ and altitude $h$ is in rotation about the base. Determine the volume of the rotating body that forms.
I've tried to find it's volume by using the formula $V = \pi \int_a^b (f(x))^2dx$, where $a = 0$, $b = 2a$ and $f(x) = kx^2$. To find $k$ I've set $f(x) = h$ and $x = 2a$ wich gives me $k = \frac{4a^2}{h}$. So I've got:
$$V = \pi \int\limits_0^{2a} \left( \frac{x^2h}{4a^2} \right)^2 dx = \frac{ h^2 \pi}{16 a^3} \int\limits_0^{2a} x^4 dx = \frac{ h^2 \pi}{16 a^3} \left( \frac{x^5}{5} \right) \bigg|_{0}^{2a} $$
And the result is totally wrong. In the answers book, the author makes a totally different thing but I didn't get what he's making. He did the follows:
$$x = \sqrt{ky}, \quad V = 2 \pi \int\limits_{a}^{h} (h - y) \sqrt{ky} dy = 2 a\pi \frac{16h^2}{15}$$
And he rotates around $x = h$. The integral itself is easy, I'm just not getting the idea behind it.
In reference to the following image:
the volume of the solid obtained by rotating the blue surface around the x-axis is equal to:
$$ V = \pi\cdot\int_{-a}^a \left(\frac{h}{a^2}\,x^2 - h\right)^2\,\text{d}x = \frac{16}{15}\,\pi\,a\,h^2\,. $$
Wanting to unnecessarily complicate life, considering this other image:
the volume of the solid obtained by rotating the blue surface around $x = h$ is equal to:
$$ V = 2\pi \cdot 2\int_0^h (h - x)\,\sqrt{\frac{a^2}{h}\,x}\,\text{d}x = \frac{16}{15}\,\pi\,a\,h^2\,. $$