]> 24.4 Volume and Surface Integrals as Single Ordinary Integrals

## 24.4 Volume and Surface Integrals as Single Ordinary Integrals

Sometimes it is possible to do volume and surface integrals as ordinary integrals without bothering with the general approach discussed in previous sections.

This happens when both your integrand and your region of integration have sufficient symmetry that you can do all but one of the integrals by inspection in the multiple integration that our formalism leads to.

Why bother with such things?

There are two reasons, neither of which is very convincing, but here they are.

First, using the general approach that applies to integrating over any reasonable shape and any reasonable integrand to solve a problem much of which you can solve by eye is like using a big gun to shoot a mosquito, or having a philosopher teach kindergarten.

Second, in traditional study of calculus you study single integration long before you study surface and volume and multiple integration, so you then have an opportunity to find answers to these questions without knowing anything about these concepts.

The absolutely simplest example is finding the area in the region between the x-axis $( y = 0 )$ two lines $x = a , x = b$ and the curve defined by the function $y = f ( x )$ .

You can write this as an area integral with integrand 1, but having done so you can immediately make this a double integral and do the integration over $y$ to get the standard formula for this area

$A = ∫ f ( x ) d x$

Suppose you rotate the curve $y = f ( x )$ about the x-axis and ask for the volume of the region generated by this action between the same limits on $x$ .

Now you can argue that the volume in a small slice of this region between $x$ and $x + d x$ is the area of the circle with radius $f ( x )$ multiplied by $d x$ . So you can write this as a single integral with integrand given by this area

$V = π ∫ f ( x ) 2 d x$

Again, we can slice our surface into sections between $x$ and $x + d x$ . The surface area in any one slice will be $d s$ multiplied by the circumference of the circle of radius $f ( x )$ .

You have to be a bit careful here because the surface sliced is generally tilted with respect to the x-axis, and the factor $d s$ here is not $d x$ but rather the length of the curve defined by $y = f ( x )$ in our slice.

We know

$d s 2 = d x 2 + d y 2$

so

$d s = d x 2 + d y 2$

or

$d s = 1 + ( d y d x ) 2 d x = 1 + f ' ( x ) 2 d x$

which gives

$S = 2 π ∫ f ( x ) 1 + f ' ( x ) 2 d x$

as the surface area of the surface generated by rotating the curve $y = f ( x )$ about the x-axis.

Similar but different formulae can be generated for regions obtained by rotating the curve defined by $y = f ( x )$ around the y-axis.

Exercise 24.6 Find single integral expressions for the volume and surface of the region generated by rotating the curve defined by $y = f ( x )$ from $( a , f ( a ) )$ to $( b , f ( b ) )$ about the y-axis.

You can also do similar things in other coordinate systems, integrating over one or more of the variables of spherical coordinates by eye, for example.

A standard example for this method is the sphere. Suppose we want to determine volume and surface area of a sphere of radius $R$ .

Notice that the sphere is a rotation of the curve defined by $x 2 + y 2 = R 2$ or $y = R 2 − x 2$ about the x-axis. The limits of integration for volume or surface are $x = − R$ and $x = R$ , and the integrands are

For volume

$π ( R 2 − x 2 )$

For surface

$2 π R 2 − x 2 1 + x 2 R 2 − x 2$

or

$2 π R$

These may be integrated with the standard results. It is mildly interesting that the surface area of a sphere in a slice of thickness $d$ is $2 π R d$ independent of where it is, as long as there are pieces of the sphere on either side of it.

Formulae like those above can be obtained for cones and wedges and all sorts of other shapes.