Volumes of Revolution

Spinning a region

Take the region under $y=\sqrt{x}$y = √x from x = 0 to 4 and spin it a full turn about the x-axis. Every point of the region sweeps a circle around the axis, and together they sweep out a solid, round in every cross-section and widening from a point at the origin to a radius of 2 at x = 4.

The region under $y=\sqrt{x}$y = √x from 0 to 4. Spun about the x-axis, the vertical strip at each x sweeps out a disc whose radius is the strip’s height.

Each slice is a disc

Cut the solid straight across the axis at x. The strip of the region there has height y, and spinning it makes a disc of radius y, so the face of the slice has area $\pi y^2$πy².

A slice of thickness dx is a thin cylinder, and a cylinder’s volume is its face area times its thickness: $\pi y^{2}dx$πy² dx.

Add the discs

Adding the slices is integrating them: $V=\pi \int y^{2}dx$V = π ∫ y² dx, with y written in terms of x and the limits in x. For $y=\sqrt{x}$y = √x from 0 to 4, $y^2=x$y² = x, and $\pi$π times the integral of x from 0 to 4 is $\pi \times \frac{16}{2}=8\pi$π × 16/2 = 8π, about 25.1 cubic units.

The radius is squared before integrating. $\pi$π times the integral of $\sqrt{x}$√x would multiply an area by $\pi$π, which is not a volume.

The line $y=1+\frac{x}{2}$y = 1 + x/2 from 0 to 4, spun about the x-axis. The strip at x = 3 has height y = 2.5 and sweeps a disc of area $\pi \times 2.5^2$π × 2.5², about 19.63. The discs from 0 to 3 add to $\pi \left(3+4.5+2.25\right)$π(3 + 4.5 + 2.25), about 30.63 cubic units. Drag the strip to where the disc has radius 2.

Checks on solids with known volumes

Spin y = x from 0 to 3 and the solid is a cone of radius 3 and height 3. $V=\pi$V = π times the integral of $x^2$x² from 0 to 3, which is $\pi \times \frac{27}{3}=9\pi$π × 27/3 = 9π: one third of the cylinder $\pi \times 3^2\times 3=27\pi$π × 3² × 3 = 27π, as the cone formula says.

Spin the semicircle $y=\sqrt{r^2-x^2}$y = √(r² − x²) from −r to r and the solid is a sphere. Here $y^2=r^2-x^2$y² = r² − x², and $\pi$π times the integral of $r^2-x^2$r² − x² from −r to r is $\pi \left(2r^3-\frac{2r^3}{3}\right)=\frac{4\pi r^3}{3}$π(2r³ − 2r³/3) = 4πr³/3, the volume of a sphere.

The whole solid in the figure, from 0 to 4, is $\pi \left(4+8+\frac{16}{3}\right)=\frac{52\pi }{3}$π(4 + 8 + 16/3) = 52π/3. It is a frustum of a cone with radii 1 and 3 and height 4, and the frustum formula $\frac{\pi h\left(R^2+R\times r+r^2\right)}{3}$πh(R² + R × r + r²)/3 gives $\pi \times 4\times \frac{13}{3}=\frac{52\pi }{3}$π × 4 × 13/3 = 52π/3 as well.

About the y-axis

Spin about the y-axis and x and y change places. The slices are cut across the y-axis, each a disc of radius x and thickness dy, so $V=\pi \int x^{2}dy$V = π ∫ x² dy, with x written in terms of y and the limits in y.

Take the region between $y=x^2$y = x², the y-axis and the line y = 4, spun about the y-axis. At height y the radius is $x=\sqrt{y}$x = √y, so $x^2=y$x² = y, and $V=\pi$V = π times the integral of y from 0 to 4, which is $\pi \times \frac{16}{2}=8\pi$π × 16/2 = 8π. The cylinder of radius 2 and height 4 around it holds $16\pi$16π, so the bowl fills half of it.

The region between $y=x^2$y = x², the y-axis and y = 4, which meet at the dot (2, 4). Spun about the y-axis, the slice at height y is a disc of radius $\sqrt{y}$√y.

The usual mistakes

Leaving the radius unsquared. A disc’s area is $\pi y^2$πy², not $\pi y$πy.

Integrating $2\pi y$2πy. That is the circumference of the disc, the edge rather than the face.

Skipping the divide. For $y=\sqrt{x}$y = √x from 0 to 4, $\pi$π times the integral of x is $\pi \times \frac{16}{2}=8\pi$π × 16/2 = 8π, not $16\pi$16π.

Taking one disc for the solid. $\pi \times 2^2=4\pi$π × 2² = 4π is the face of the last slice only.

Keeping dx about the y-axis. The slices are stacked up the y-axis, so the radius is x and the thickness dy.

A turned vase

In the application below, a vase is turned on a lathe about a vertical axis, so its slices are cut across that axis and stacked in dy. Each slice of the cavity is a disc; each slice of the wall is a ring, a disc with a smaller disc taken out.