### Volumes by cross sections

Next to computing areas of regions in the plane, the easiest *concept* of application of the ideas of calculus is to computing
volumes of solids where somehow we know a formula for the *areas
of slices*, that is, *areas of cross sections*. Of course, in any
particular example, the actual issue of getting the formula for the
cross section, and figuring out the appropriate limits of integration,
can be difficult.

The idea is to
just ‘add them up’:
$$\hbox{ volume }=\int_{\hbox{left limit}}^{\hbox{right limit}}
\hbox{ (area of cross section at $x$) }\;dx$$
where in whatever manner we describe the solid it extends from
$x=$*left limit* to $x=$*right limit*. We must suppose
that we have some reasonable *formula* for the area of the cross
section.

For example, let's find the volume of a solid ball of radius $1$. (In
effect, we'll be deriving the formula for this). We can
suppose that the ball is centered at the origin. Since the radius is
$1$, the range of $x$ coordinates is from $-1$ to $+1$, so $x$ will be
integrated from $-1$ to $+1$. At a particular value of $x$, what does
the cross section look like? A disk, whose radius we'll have to
determine. To determine this radius, look at how the solid ball
intersects the $x,y$-plane: it intesects in the disk $x^2+y^2\le
1$. For a particular value of $x$, the values of $y$ are between
$\pm\sqrt{1-x^2}$. This line segment, having $x$ fixed and $y$ in this
range, is the intersection of the cross section disk with the
$x,y$-plane, and in fact is a *diameter* of that cross section
disk. Therefore, the radius of the cross section disk at $x$ is
$\sqrt{1-x^2}$. Use the formula that the area of a disk of radius $r$
is $\pi r^2$: the area of the cross section is
$$\hbox{ cross section at $x$ } = \pi(\sqrt{1-x^2})^2 = \pi(1-x^2) $$
Then integrate this from $-1$ to $+1$ to get the volume:
\begin{align*}
\hbox{ volume } &= \int_{\rm left}^{\rm right}\,\hbox{area of
cross-section}\,dx\\
&= \int_{-1}^{+1}\, \pi(1-x^2)\,dx = \pi[x-{x^3\over 3}]_{-1}^{+1}\\
&= \pi[(1-{1\over 3}) - (-1-{(-1)^3 \over 3})] = {2\over 3} + {2\over
3}
= {4\over 3}
\end{align*}

#### Exercises

- Find the volume of a circular cone of radius $10$ and height $12$ (not by a formula, but by cross sections).
- Find the volume of a cone whose base is a
*square*of side $5$ and whose height is $6$, by cross-sections. - A hole $3$ units in radius is drilled out along a diameter of a solid sphere of radius $5$ units. What is the volume of the remaining solid?
- A solid whose base is a disc of radius $3$ has
vertical cross sections which are
*squares*. What is the volume?

#### Thread navigation

##### Calculus Refresher

- Previous: Centers of mass (centroids)
- Next: Volume of surfaces of revolution