Question

Calculate the volume of the solid formed by revolving about the line y=1 the region bounded by the parabola

𝑥

2 = 4𝑦 and that line. Take the rectangular elements of area parallel to the axis of revolution.

Answer 1

The length of the square area element for fixed y is

$\sqrt{4y}-\left( -\sqrt{4y} \right) =4\sqrt{y}$

Then the volume of this element rotated is

$dV=2\pi \left( 1-y \right) \cdot 4\sqrt{y}dy$ .

Then

$V=\int_0^1{2\pi \left( 1-y \right) \cdot 4\sqrt{y}dy}=8\pi \int_0^1{\left( y^{1/2}-y^{3/2} \right) dy}=\\=8\pi \left( \frac{2}{3}-\frac{2}{5} \right) =\frac{32\pi}{15}$