Volume

** This material is section 6.2 in the textbook**

Integration finds the area under a curve, or between curves, by sweeping out that area with a line segment. That is, integration finds a 2 dimensional ‘amount’ by breaking it into 1-dimensional slices, and adding them up. The same trick works for volumes!

The volume of a shape can be calculated by integration, by integrating the area of a collection of cross sections.

Volume of a cylinder: area of a slice = $\pi r^2$: $$V=\int_0^h \pi r^2dx=\pi r^2 h$$

Volume of a cone: Say cone has base with radius $r$ and height $h$: Area of a cross section $\pi r(x)^2$ where $r(x)$ is the radius at height $x$: This radius goes from $0$ to $r$ over a distance of $h$: so it has slope $r/h$. Thus = $r/h x$

$$Volume = \int_0^h A(x) dx=\int_0^h \pi (r/h x)^2 dx=\frac{\pi r^2}{h^2}\int_0^h x^2 dx$$ $$=\pi r^2\frac{h^3/3}{h^2}=\pi r^2 h/3$$

(Note: you could set this up in a variety of ways: if you put the cone the other direction you’d get a radius function of $r-r/h x$ and it’ll integrate to the same answer!)

Volume of a square pyramid: base has side length $s$, and height $h$: Cross sections are squares, so area is $l(x)^2$ where $l$ is the side length of the square. This side length goes from $0$ to $s$ as $x$ goes from $0$ to $h$, and does so linearly, so the length at $x$ is $l(x)=s/h x$. Thus the volume is

$$\int_0^h (s/h x)^2 dx=\frac{s^2}{h^2}\int_0^h x^2dx=s^2h/3$$

Volume of a sphere: Cross sections are circles, so radius is $\pi r(x)^2$. What is the radius of a cross section at height $x$? $\sqrt{r^2-x^2}$: Thus, $$Vol = \int_{-r}^r A(x)dx=\int_{-r}^r \pi (r^2-x^2)dx=\frac{4}{3}\pi r^3$$

How easy it is to calculate the volume depends on how you slice it: Slicing a cube horizontally gives square cross sections, but slicing diagonally gives triangles which turn into hexagons, then back to triangles.

What about the volume outside of a cone in a cylinder? We could find the volume of the cone and subtract it, so we already know the answer. But let’s try to calculate it directly, where the cross sections are ring shapes. These rings are a disk with a disk punched out of them, so their area is $$A=Outer-Inner$$ If $R$ is the radius of the outer disk and $r$ the radius of the inner disk, $$V=\int \pi R^2-\pi r^2$$

All of this can be made rather systematic for volumes of revolution. A volume of revolution is the rotation of some function $f(x)$ in the plane about a line (either x or y axis usually, sometimes another line). The cross sections of such a shape will always be circles, so the integral will always look like $$\int Area ds =\int \pi (radius)^2 ds$$ Where $s$ is the slicing variable: what does the slice depend on?

Example: Find the volume inside rotating $\sqrt(x)$ around the $x$ axis from 0 to 1.

Find the volume from rotating $\sqrt(x)$ around the $y$ axis from $0$ to 1.

Set up but don’t evaluate the following volumes:

Rotate $y=\ln(x),y=0$ about the $x$ axis from $x=1$ to $x=3$. Rotate $y=\ln(x)$ about the $y$ axis from $x=0$ to $x=1$.

Rotate 2x=y^2, x=0,y=4$ about the $y-$axis.

Volumes using rings (washers)

If you rotate the volume bounded by two curves about an axis, the shape traced out will be composed of rotated rings, instead of disks. These are just disks, with a smaller disk punched out of the middle. So, instead of finding the (single) radius defining the disk, you need an inner and outer radius:

Find volume cut out by rotating region bounded by $y=x, y=x^2$ about the $x$ axis. Find volume cut out by revolving same region about the $y$ axis.

Set up the integral for the volume bounded by $3-x^2$ and $x=1$ about the $x$ axis.

Volumes using cylindrical shells

Cut a rotated volume up by slices parallel to the rotation axis leaves cylindrical shells. The area of a cylinder is its circumference times its height, or $2\pi rh$. Thus, the volume is $$\int Area dx = \int 2\pi r h dx$$ and the goal is to find $r$ and $h$ from the function $f$ being rotated: