The volume of a solid of revolution

The volume of a solid made by turning a curve about an axis is also found with a definite integral¹. We revolve the part of $y = \sqrt{x}$ with $0 \leq x \leq 4$ about the $x$ -axis.

Adding up thin discs

Cutting perpendicular to the axis at $x$ gives a circular section of radius $f(x)$ and area $\pi f(x)^2$ . Adding them up with thickness $dx$ gives the volume.

V = \pi \int_a^b f(x)^2\,dx

Adding cross-sectional areas is one level above adding vertical segments, which is what gave an area.

Working it out

For $f(x) = \sqrt{x}$ we have $f(x)^2 = x$ , so the computation is easy.

V = \pi \int_0^4 x\,dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = 8\pi \approx 25.13

The square root disappears under the square, so the volume of a revolved parabola comes from integrating a linear function.

Compared with a cylinder

Solid	Volume
The solid of revolution	$8\pi$
The enclosing cylinder, radius $2$ and height $4$	$16\pi$
Ratio	$\dfrac{1}{2}$

Since $f(x)^2 = x$ is linear, the cross-sectional area grows from $0$ at a constant rate, which is what the ratio reflects.

Changing the axis

To revolve about the $y$ -axis, solve for $x$ first and use the same formula. Revolving $x = y^2$ over $0 \leq y \leq 2$ gives the following.

V = \pi \int_0^2 y^4\,dy = \frac{32\pi}{5}

Change the axis and the volume changes.

When there is a hole

Revolving the region between two curves $f$ and $g$ with $f \geq g$ , subtract the inner from the outer.

V = \pi \int \left( f^2 - g^2 \right) dx

It is the same shape as subtracting the lower function from the upper one for an area. The computation of a parabolic dish or the capacity of a vessel is written this way.

The two curves on the graph are $y = \pm\sqrt{x}$ , the outline before revolving. The large dots are the rims at the end of the solid.

Solid of revolution, Wikipedia