It is possible to find the volume of a solid using Integration and the Fundamental Theorem of Calculus. There are two methods, known as the disc and shell methods, both of which involve the summation of much smaller solids to create the whole solid. The volume of a solid with a given cross section can also be found using a similar method. The calculus enters in order to find the sum of the infinitely thin discs, shells and cross-sections.

Creating the Solid

In many instances, the difficulty of the problem lies not with the calculus (which is often just a definite integral over given bounds) but with actually visualising the solid. The solid is bound by one or more functions over a given interval, and an axis parallel to either the X or Y axis.
If it is a rotational solid, it is then rotated around the axis, forming a 3D object coming out of the front and back of the page.
(Graph of a function)
If the solid is made up of cross-sections, the solid is all of the cross-sections placed together so that they fit within the bounds. The cross sections will run parallel to one of the axes and change in size as the values of X and Y change.
(Graph of a function)

Disc Method

Using the disk method to find the volume of a rotational solid, we use a number of thin disks to approximate the volume of the solid. A disk (a solid but thin cylinder) has the dimensions (the area of a circle) multiplied by h, the height. However, for our purposes, h will be replaced by either a or a , representing the fact that it is a small slice of the function. This will depend on whether the disc is vertical (when the h will be ) or horizontal (when h will be ). As a result a disc for the function would look like:

If we want a better approximation, it makes sense that we make the disc increasingly thinner, much like in a Reiman Sum. Our approximation gets better as the h value approaches infinite thinness, and equal to dx or dy. Now we can add all of the infinitely thin slices together using the Fundamental Theory of Calculus.

The standard formula for the volume of a disc with thickness dx looks like this:

And the volume for the entire solid on the interval [a,b] looks like this:

If the discs are stacked horizontally, the height will be dy instead of dx.

Shell Method

The shell in the shell method is actually a hollow cylinder. Finding the volume, however, is less straightforward than the volume of a disc. It has a a height h, and a thickness dx or dy, but the fact that the cylinder is hollow presents a problem. The way we overcome this is by "opening up" the cylinder to make it a flat sheet. We can only do this because the shell "wall" is infinitely thin, so the difference between the volume of the shell and the sheet is very small. The circumference (the rim of the shell) is equal to , so when we open the function, the length of the sheet is also . r is the distance from the axis of revolution to the shell, and will therefore be the value of x or y for the function. Using the function from above, the shell and the sheet look like this:

Cross Sections

When asked to find the volume of a solid with a given cross section, the volume of the solid is broken into slices, which we then integrate to form the whole. The slices have a thickness or which is multiplied by the area of the cross section. For instance, if the cross section is a square with a side length l, the volume will be . The length of the side will either be the x or y value of the function.

Solids With More Than One Function

Often you will be faced with a problem involving a solid bound by more than one function. In this case, it will often be necessary to find the bounds of the integral by setting the two equations equal to each other and solving. When using the disc method, the volume will be that of the bigger disc minus the volume of the smaller disc, which are then all added up in the integral (usually two separate integrals for simplicity). When using the shell method, the shell's height will be the function with a greater y value (i.e. the top function) minus the function with the lower y value (the one on the bottom). For cross sections, the base on the x-y plane has the value of the greater function minus the lesser function. In all cases the objective is to find the length of the slice actually added up to find the area.

## Volume of Solids

## Table of Contents

## Creating the Solid

In many instances, the difficulty of the problem lies not with the calculus (which is often just a definite integral over given bounds) but with actually visualising the solid. The solid is bound by one or more functions over a given interval, and an axis parallel to either the X or Y axis.

If it is a rotational solid, it is then rotated around the axis, forming a 3D object coming out of the front and back of the page.

(Graph of a function)

If the solid is made up of cross-sections, the solid is all of the cross-sections placed together so that they fit within the bounds. The cross sections will run parallel to one of the axes and change in size as the values of X and Y change.

(Graph of a function)

## Disc Method

If we want a better approximation, it makes sense that we make the disc increasingly thinner, much like in a Reiman Sum. Our approximation gets better as the h value approaches infinite thinness, and equal to

dxordy. Now we can add all of the infinitely thin slices together using the Fundamental Theory of Calculus.The standard formula for the volume of a disc with thickness

dxlooks like this:And the volume for the entire solid on the interval [a,b] looks like this:

If the discs are stacked horizontally, the height will be

dyinstead ofdx.## Shell Method

The shell in the shell method is actually a hollow cylinder. Finding the volume, however, is less straightforward than the volume of a disc. It has a a height h, and a thickness

dxordy, but the fact that the cylinder is hollow presents a problem. The way we overcome this is by "opening up" the cylinder to make it a flat sheet. We can only do this because the shell "wall" is infinitely thin, so the difference between the volume of the shell and the sheet is very small. The circumference (the rim of the shell) is equal to , so when we open the function, the length of the sheet is also .ris the distance from the axis of revolution to the shell, and will therefore be the value ofxoryfor the function. Using the function from above, the shell and the sheet look like this:## Cross Sections

When asked to find the volume of a solid with a given cross section, the volume of the solid is broken into slices, which we then integrate to form the whole. The slices have a thickness or which is multiplied by the area of the cross section. For instance, if the cross section is a square with a side length

l, the volume will be . The length of the side will either be thexoryvalue of the function.## Solids With More Than One Function

Often you will be faced with a problem involving a solid bound by more than one function. In this case, it will often be necessary to find the bounds of the integral by setting the two equations equal to each other and solving. When using the disc method, the volume will be that of the bigger disc minus the volume of the smaller disc, which are then all added up in the integral (usually two separate integrals for simplicity). When using the shell method, the shell's height will be the function with a greater y value (i.e. the top function) minus the function with the lower y value (the one on the bottom). For cross sections, the base on the x-y plane has the value of the greater function minus the lesser function. In all cases the objective is to find the length of the slice actually added up to find the area.

## Practice Problems

## Additional Resources