Volumes of Revolution by Slicing

Start with an area --- a planar region --- which you can imagine as a piece of cardboard. The cardboard is attached by one edge to a stick (the axis of revolution). As you spin the stick, the area revolves and sweeps out a region in space.

$$\hbox{\epsfysize=2in \epsffile{vrev1.eps}}$$

The problem is to find the volume of revolution --- the volume of the region in space which is swept out by the area.

In the simplest case --- the one I've described above --- you can find the volume by cutting the region into circular slices perpendicular to the axis.

$$\hbox{\epsfysize=2in \epsffile{vrev2.eps}}$$

If the radius of a slice is r, then its volume is

$$\pi r^2\cdot (\hbox{thickness}).$$

In the simplest case, the radius is given by a nonegative function $r = f(x)$ , and the volume is generated by revolving the area under the graph of f from $x = a$ to $x = b$ around the x-axis. As usual, I divide the interval $a \le x \le b$ up into pieces, with the k-th piece having width $\Delta x_k$ . I pick an x-value in the k-th piece, say $x_k$ . Then the volume of the k-th circular slice will be

$$\pi f(x_k)^2 \Delta x_k.$$

The total volume is approximated by adding up the volumes of the slices, as you can see in the picture above. So if I have n slices, then

$$(\hbox{total volume}) \approx \sum_{k=1}^n \pi f(x_k)^2 \Delta x_k.$$

To get the exact volume, I shrink the slices, letting the thickness $\Delta x_k$ of a typical slice go to 0:

$$(\hbox{total volume}) = \lim_{\Delta x_k} \sum_{k=1}^n \pi f(x_k)^2 \Delta x_k.$$

The expression on the right is a Riemann sum for $\displaystyle \int_a^b \pi f(x)^2\,dx$ . So

$$(\hbox{total volume}) = \int_a^b \pi f(x)^2\,dx.$$

In setting up problems, I'll use a shortcut rather than writing down the Riemann sum. I'll simply write

$$(\hbox{total volume}) = \int_a^b \pi r^2\cdot (\hbox{thickness}).$$

In the examples I do, the axis of revolution will always be parallel to the x-axis or the y-axis. If the axis is parallel to the x-axis, the thickness is $dx$ ; if the axis is parallel to the y-axis, the thickness is $dy$ .

The radius r is the distance from the axis of revolution to the edge of the slice. It will usually be given by a function specified in the problem which determines the region which is being revolved. You'll see how this works in the examples below.

Example. Here is the region under $y = \sin x$ from $x = 0$ to $x = \pi$ :

$$\hbox{\epsfysize=1.5in \epsffile{vrev3.eps}}$$

The region is revolved about the x-axis. It sweeps out a volume of revolution:

$$\hbox{\epsfysize=1.75in \epsffile{vrev4.eps}}$$

To find the volume, cut the solid into circular slices --- like sausage slices --- perpendicular to the axis:

$$\hbox{\epsfysize=1.75in \epsffile{vrev5.eps}}$$

The radius of a typical slice is the height of the curve: $r = \sin x$ . The thickness is $dx$ . Thus, the volume of a typical slice is $\pi(\sin x)^2\,dx$ --- (circle area) times (thickness). So the total volume is

$$V = \int_0^\pi \pi(\sin x)^2\,dx = \pi \int_0^\pi \dfrac{1}{2}(1 - \cos 2x)\,dx = \dfrac{\pi}{2}\left[x - \dfrac{1}{2}\sin 2x\right]_0^\pi = \dfrac{\pi^2}{2}.$$

Example. The area bounded by $y = 6x - x^2$ and the x-axis is revolved about the x-axis. Find the volume of the solid generated.

The region is the area under the parabola $y = 6x - x^2$ from $x = 0$ to $x = 6$ .

$$\hbox{\epsfysize=1.75in \epsffile{vrev6.eps}}$$

In the picture above, I've superimposed a typical slice over the picture of the area being revolved. You can see that the radius of a typical slice is the height of the curve: $r = 6x - x^2$ . The thickness of a typical slice is $dx$ .

Thus, the total volume is

$$V = \int_0^6 \pi(6x - x^2)^2\,dx = \pi \int_0^6 (36x^2 - 12x^3 + x^4)\,dx = \pi \left[12x^3 - 3x^4 + \dfrac{1}{5}x^5\right]_0^6 = \dfrac{1296\pi}{5}.\quad\halmos$$

In the situations above, the axis of revolution lay along one edge of the area. If it does not, the volume of revolution may have a "hole" in the middle:

$$\hbox{\epsfysize=1.75in \epsffile{vrev7.eps}}$$

If I cut such a volume into slices, I get circular rings or washers:

$$\hbox{\epsfysize=1.75in \epsffile{vrev8.eps}}$$

The area of such a washer is

$$\pi r_{\rm out}^2 - \pi r_{\rm in}^2 = \pi(r_{\rm out}^2 - r_{\rm in}^2).$$

Thus, the volume of a typical washer is

$$\pi(r_{\rm out}^2 - r_{\rm in}^2)\cdot (\hbox{thickness}).$$

As before, I integrate to find the total volume:

$$V = \int_a^b \pi(r_{\rm out}^2 - r_{\rm in}^2)\cdot (\hbox{thickness}).$$

Example. The region bounded by $y = x^2$ and $y = 1$ is revolved about the x-axis. Find the volume of the solid that is generated.

$$\hbox{\epsfysize=1.75in \epsffile{vrev9.eps}}$$

For a typical washer, the inner radius is $r_{\rm in} = x^2$ and the outer radius is $r_{\rm out} = 1$ . You can see this in the picture above, where I've superimposed a typical washer on top of the picture of the region.

The region extends from $x = -1$ to $x = 1$ . The volume is

$$V = \int_{-1}^1 \pi(1^2 - (x^2)^2)\,dx = \pi \int_{-1}^1 (1 - x^4)\,dx = \pi\left[x - \dfrac{1}{5}x^5\right]_{-1}^1 = \dfrac{8\pi}{5}. \quad\halmos$$

Example. The region bounded by $x = y^2$ and $x = 2 - y$ is revolved about the y-axis. Find the volume generated.

$$\hbox{\epsfysize=1.75in \epsffile{vrev10.eps}}$$

The curves intersect at $y = -2$ and at $y = 1$ . The inner radius is $r_{\rm in} = y^2$ and the outer radius is $r_{\rm out} = 2 - y$ . The volume of the solid is

$$V = \int_{-2}^1 \pi\left((2 - y)^2 - (y^2)^2\right)\,dy = \pi\left[-\dfrac{1}{3}(2 - y)^3 - \dfrac{1}{5}y^5\right]_{-2}^1 = \dfrac{72\pi}{5}.\quad\halmos$$

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Last updated: December 2, 2005

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