8.4: Surfaces and Solids of Revolution

Long before calculus was invented the ancient Greeks (e.g. Archimedes) discovered the formulas for the volume and surface area of familiar three-dimensional objects such as the sphere.⁶ Volumes and surface areas of arbitrary solids and surfaces can be found using multivariable calculus. However, single-variable calculus can be used in the special case of the objects possessing symmetry about an axis, via methods that involve revolving a curve or region in the \(xy\)-plane around an axis.

For example, revolve a curve \(y=f(x) \ge 0\) around the \(x\)-axis, for \(a \le x \le b\). This produces a surface of revolution in three dimensions, as in Figure [fig:surfarea](a).

To find the total lateral surface area \(S\), pick \(x\) in \(\lival{a}{b}\) then find the infinitesimal surface area \(d\!S\) swept out over the infinitesimal interval \(\ival{x}{x+\dx}\), as in Figure [fig:surfarea](b). By the Microstraightness Property, the curve \(y=f(x)\) is a straight line segment of length \(\ds\) over that interval, so that the infinitesimal surface is a frustrum—a right circular cone with the vertex chopped off by a plane parallel to the base circle. From geometry⁷ you might recall the formula for the lateral surface area of the frustrum in Figure [fig:frustrum]: \(\pi\,(r_1+r_2)\,l\). Use that formula with \(r_1=f(x)\), \(r_2=f(x+\dx)=f(x)+\dy\), and \(l=\ds=\sqrt{1 + (f'(x))^2}\,\dx\) (by formula ([eqn:arclength]) in Section 8.3) as in Figure [fig:surfarea](c), so that \(d\!S\) is

\[\begin{aligned} d\!S ~&=~ \pi\,(f(x) + (f(x)+\dy))\,\sqrt{1 + (f'(x))^2}\,\dx\\ &=~ 2\,\pi\,f(x)\,\sqrt{1 + (f'(x))^2}\,\dx ~+~ \pi\,\sqrt{1 + (f'(x))^2}\,\dy\,\dx\\ &=~ 2\,\pi\,f(x)\,\sqrt{1 + (f'(x))^2}\,\dx ~+~ 0\end{aligned} \nonumber \]

since \(\dy\,\dx = f'(x)\,(\dx)^2 = 0\). The surface area \(S\) is then the sum of all the areas \(d\!S\):

Note that formula ([eqn:surfareagen]) holds by symmetry and formula ([eqn:surfarea]). A curve \(y=f(x) <0\) and the curve \(y=\Abs{f(x)}=-f(x)\) are symmetric with respect to the \(x\)-axis, as in the figure on the right. Thus, both curves sweep out the same surface of revolution when revolved around the \(x\)-axis. This means formula ([eqn:surfareagen]) also holds if \(y=f(x)\) changes sign in \(\ival{a}{b}\): similar to the area between two curves, you would split the integral over different subintervals depending on the sign.

Show that the surface area of a sphere of radius \(r\) is \(4\pi r^2\).

Solution: Use the circle \(x^2+y^2=r^2\). The upper half of that circle is the curve \(y=f(x)=\sqrt{r^2-x^2}\) over the interval \(\ival{-r}{r}\), as in the figure on the right. Revolving that curve around the \(x\)-axis produces a sphere of radius \(r\), whose surface area \(S\) is:

\[\begin{aligned} S ~&=~ \int_{-r}^r 2\,\pi\,f(x)\,\sqrt{1 + (f'(x))^2}~\dx\\ &=~ \int_{-r}^r 2\,\pi\,\sqrt{r^2-x^2}\,\sqrt{1 + \left(\frac{-x}{\sqrt{r^2-x^2}}\right)^2}~\dx\\ &=~ \int_{-r}^r 2\,\pi\,\cancel{\sqrt{r^2-x^2}}\,\sqrt{\frac{r^2}{\cancel{r^2-x^2}}}~\dx\\ &=~ 2\,\pi\,rx~\Biggr|_{-r}^r ~=~ 4\,\pi r^2 \quad\checkmark\end{aligned} \nonumber \]

A similar derivation using a frustrum yields the surface area \(S\) of the surface of revolution obtained by revolving a curve \(y=f(x)\) around the \(y\)-axis, for \(0\le a\le x\le b\):

\[\label{eqn:surfareageny} S ~=~ \int_a^b d\!S ~=~ \int_a^b 2\,\pi\,\abs{x}~\ds ~=~ \int_a^b 2\,\pi\,x\,\sqrt{1 + (f'(x))^2}~\dx \]

Now suppose you revolve the region between a curve \(y=f(x)\ge 0\) and the \(x\)-axis around the \(x\)-axis, for \(a\le x\le b\) (see Figure [fig:solidvolume](a)). This produces a solid of revolution in three dimensions, as in Figure [fig:solidvolume](b). Notice that this solid consists of the surface of revolution as before along with its interior.

The goal is to find the volume \(V\) of this solid. The idea is to divide the solid into slices, like a loaf of bread. First, the infinitesimal volume \(d\!V\) of the frustrum swept out by a strip of infinitesimal width \(\dx\) at \(x\) in \(\lival{a}{b}\)—shown in Figure [fig:solidvolume](c)—is needed. By the Microstraightness Property the curve \(y=f(x)\) is a straight line of length \(\ds\) over the interval \(\ival{x}{x+\dx}\). There is thus a right triangle at the top of the strip—unshaded in Figure [fig:solidvolume](c)—whose area \(A\) is zero: \(A = \tfrac{1}{2} (\dy)(\dx) = \tfrac{1}{2} f'(x) (\dx)^2 = 0\).

That triangle thus contributes no volume when revolved around the \(x\)-axis: the volume \(d\!V\) swept out by that strip all comes from the shaded rectangle of height \(f(x)\) and width \(\dx\). That rectangle sweeps out a right circular cylinder of radius \(f(x)\) and height \(\dx\) (see Figure [fig:discmethod]). The volume of a right circular cylinder of radius \(r\) and height \(h\) is defined as the area of the base circle times the height: \(\pi r^2h\). Hence,

\[d\!V ~=~ \pi\,(f(x))^2\;\dx ~~. \nonumber \]

The total volume \(V\) of the solid is then the sum of all those infinitesimal volumes \(d\!V\):

This method for finding the volume is called the disc method, since the cylinder of volume \(d\!V\) resembles a disc. Think of the discs as being similar to infinitesimally thin slices of a loaf of bread. Notice that absolute values are not needed in formula ([eqn:volumedisc]) since \(f(x)\) is squared, so the formula holds even when \(f(x)\) is negative.

Show that the volume of a sphere of radius \(r\) is \(\frac{4}{3} \pi r^3\).

Solution: Use the circle \(x^2+y^2=r^2\). Revolve the region between the upper half of the circle \(y=f(x)=\sqrt{r^2-x^2}\) and the \(x\)-axis around the \(x\)-axis over the interval \(\ival{-r}{r}\), as in the figure on the right. The solid of revolution swept out is a sphere of radius \(r\), whose volume \(V\) is:

\[\begin{aligned} V ~&=~ \int_{-r}^r \pi\,(f(x))^2~\dx ~=~ \int_{-r}^r \pi\,(r^2-x^2)~\dx ~=~ \pi\,r^2 x ~-~ \frac{1}{3} \pi x^3~\Biggr|_{-r}^r\\ &=~ \left(\pi\,r^3 - \frac{1}{3} \pi r^3\right) ~-~ \left(-\pi\,r^3 + \frac{1}{3} \pi r^3\right) ~=~ \frac{4}{3}\,\pi r^3 \quad\checkmark\end{aligned} \nonumber \]

Instead of memorizing formula ([eqn:volumedisc]), try to remember the more generic approach of revolving an infinitesimal rectangular strip around an axis, which might not be the \(x\)-axis. The idea is to find the radius \(r\) and height \(h\)—typically \(\dx\) or \(\dy\)—of the disc swept out by that strip, so that the disc’s volume is \(d\!V = \pi r^2 h\). Then integrate \(d\!V\) over the appropriate interval to find the volume \(V\) of the entire solid.

Suppose the region bounded by the curve \(y=x^2\) and the \(x\)-axis for \(0 \le x \le 1\) is revolved around the line \(x=1\). Find the volume of the resulting solid of revolution.

Solution: The region is shaded in the figure on the right. Since the region is revolved around a vertical axis, the disc method will use discs with height \(\dy\), not \(\dx\). At a point \(x\) in \(\ival{0}{1}\) go up to the curve \(y=x^2\) and draw a horizontal rectangular strip to the line \(x=1\), as shown in the figure. Let \(h=\dy\) and revolve that strip around the line \(x=1\), producing a disc of radius \(r-1-x\) and height \(h=\dy\). Since \(y=x^2\) implies \(x=\sqrt{y}\), the volume \(d\!V\) of that disc is

\[d\!V ~=~ \pi\,r^2 h ~=~ \pi\,(1-x)^2~\dy ~=~ \pi\,(1-\sqrt{y})^2~\dy ~=~ \pi\,(1 - 2\sqrt{y} + y)~\dy ~. \nonumber \]

The volume \(V\) of the entire solid is then the sum of those volumes \(d\!V\) along the \(y\)-axis for \(0\le y\le 1\):

\[V ~=~ \int_0^1 d\!V ~=~ \int_0^1 \pi\,(1 - 2\sqrt{y} + y)~\dy ~=~ \pi\,\left(y - \frac{4}{3}y^{3/2} + \frac{1}{2}y^2 \right)~\Biggr|_0^1 ~=~ \pi\,\left(1 - \frac{4}{3} + \frac{1}{2}\right) ~=~ \frac{\pi}{6} \nonumber \]

The shell method can be used for finding the volume of a solid with a “hole” in the middle, as in the solid of revolution produced by revolving the shaded region in the figure on the right around the \(y\)-axis. The hole in the solid between \(x=-a\) and \(x=a\) is a result of the gap between the \(y\)-axis and the region. To find the volume \(V\) of that solid, at a point \(x\) in \(\lival{a}{b}\) form an infinitesimal strip of width \(\dx\) from the \(x\)-axis up to the curve \(y=f(x)\), as in Figure [fig:shellmethod](a).

Just like the strip in the disc method, the right triangle at the top of this strip—as in Figure [fig:shellmethod](b)—has zero area and thus does not contribute to the volume \(d\!V\) of the right circular cylindrical shell swept out by the strip, shown in Figure [fig:shellmethod](c). The volume of that shell is just the volume of the “outer” cylinder of radius \(x+\dx\) minus the volume of the “inner” cylinder of radius \(x\), both with height \(f(x)\):

\[\begin{aligned} d\!V ~&=~ \pi\,(x+\dx)^2\,f(x) ~-~ \pi\,x^2\,f(x)\

\[-6pt] &=~ \cancel{\pi\,x^2\,f(x)} ~+~ 2\,\pi\,x\,f(x)\,\dx ~+~ \pi\,\cancelto{0}{(\dx)^2}\,f(x) ~-~ \cancel{\pi\,x^2\,f(x)}\\ &=~ 2\,\pi\,x\,f(x)\,\dx\end{aligned} \nonumber \]

The volume \(V\) of the entire solid is then the sum of those volumes \(d\!V\), using an absolute value to handle any sign for \(f(x)\):

Suppose the region bounded by the curve \(y=x^2\) and the \(x\)-axis for \(0 \le x \le 1\) is revolved around the \(y\)-axis. Find the volume of the resulting solid of revolution.

Solution: The region is shaded in the figure on the right. The vertical strip at \(x\) in \(\lival{0}{1}\) with infinitesimal width \(\dx\) and height \(\Abs{f(x)}=f(x)\) is shown in the figure. That strip produces the shell with volume \(d\!V\) in formula ([eqn:volumeshell]), so by the shell method the volume \(V\) of the solid of revolution is:

\[V ~=~ \int_0^1 d\!V ~=~ \int_0^1 2\,\pi\,x\,\Abs{f(x)}~\dx ~=~ \int_0^1 2\,\pi\,x\,\cdot\,x^2~\dx ~=~ \frac{\pi}{2}\,x^4~\Biggr|_0^1 ~=~ \frac{\pi}{2} \nonumber \]

The volume \(d\!V\) in formula ([eqn:volumeshell]) can be generalized to \(d\!V = 2 \pi r h w\), where \(r\) is the distance from the axis of revolution to a generic vertical strip of infinitesimal width \(w\) in the region, and \(h\) is the height of the strip.

Example \(\PageIndex{1}\): shell1

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Solution

Suppose the region bounded by the curves \(y=x^2\) and \(y=x\) is revolved around the \(y\)-axis. Find the volume of the resulting solid of revolution.

Solution: The region is shaded in the figure on the right, along with a vertical strip with infinitesimal width \(w=\dx\) at the distance \(r=x\) from the \(y\)-axis in the region and height \(h=x-x^2\). That strip produces the shell with volume \(d\!V=2\pi rhw=2\pi x(x-x^2)\,\dx\), so that the volume \(V\) of the solid of revolution is:

\[V ~=~ \int_0^1 d\!V ~=~ \int_0^1 2\,\pi\,x\,(x-x^2)~\dx ~=~ \frac{2\pi}{3}\,x^3 ~-~ \frac{\pi}{2}\,x^4~\Biggr|_0^1 ~=~ \frac{\pi}{6} \nonumber \]

[sec8dot4]

For Exercises 1-3, find the surface area of the surface of revolution produced by revolving the given curve around the \(x\)-axis for the given interval.

3

\(y = \sqrt{4 - x^2}~\) for \(~1 \le x \le 2\vphantom{\frac{x^3}{6}}\)

\(y = \cosh\,x~\) for \(~0 \le x \le 1\vphantom{\frac{x^3}{6}}\)

\(y = \frac{x^3}{6} + \frac{1}{2x}~\) for \(~1 \le x \le 3\)

For Exercises 4-6, find the volume of the solid of revolution produced by revolving the region between the given curve and the \(x\)-axis around the \(x\)-axis for the given interval. [[1.]]

3

\(y = x^3~\) for \(~0 \le x \le 1\)

\(y = \sin\,x~\) for \(~0 \le x \le \pi\)

\(y = \sqrt{x}~\) for \(~0 \le x \le 1\)

For Exercises 7-9, find the volume of the solid of revolution produced by revolving the region between the given curve and the \(x\)-axis around the \(y\)-axis for the given interval. [[1.]]

3

\(y = \sin\,(x^2)~\) for \(~0 \le x \le \sqrt{\pi}\)

\(y = \sin\,x~\) for \(~0 \le x \le \pi\)

\(y = x^2 - x^3~\) for \(~0 \le x \le 1\)

Revolve the region in Example

Example \(\PageIndex{1}\): shell1

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Solution

around the line \(x=1\) and find the volume of the resulting solid.

[exer:ellipsoid] Revolving the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) around the \(x\)-axis produces an ellipsoid, for \(a>b>0\). Show that the surface area of the ellipsoid is \(2\,\pi\,b^2\,\left(1 + \frac{a}{eb}\,\sin^{-1} e\right)\), where \(e\) is the eccentricity of the ellipse.

Show that the volume inside the ellipsoid from Exercise [exer:ellipsoid] is \(\frac{4}{3}\,\pi\,a\,b^2\).

Find the surface area and volume of a right circular cone of radius \(r\) and height \(h\).

Formulas ([eqn:surfareagen]), ([eqn:volumedisc]) and ([eqn:volumeshell]) can be extended to include regions over infinite intervals—the integrals in those formulas simply become improper integrals. Consider the region between the curve \(y=\frac{1}{x}\) and the \(x\)-axis over the interval \(\lival{1}{\infty}\). Revolve that region around the \(x\)-axis.

1. Show that the surface area of the resulting surface of revolution is infinite.
2. Show that the volume of the resulting solid of revolution is \(\pi\).

[exer:torus] For \(0<a<b\), revolving the region inside the circle \((x-b)^2+y^2=a^2\) around the \(y\)-axis produces a donut-shaped solid of revolution called a torus. Show that the volume of the torus is \(2\pi^2a^2b\).

Use formula ([eqn:surfareageny]) and symmetry to show that the torus from Exercise [exer:torus] has surface area \(4\pi^2ab\).