First, just what is a torus? A torus is a donut shaped solid that is generated by rotating the circle of radius \(r\) and centered at (\(R\), 0) about the \(y\)-axis. This is shown in the sketch to the left below.

One of the trickiest parts of this problem is seeing what the cross-sectional area needs to be. There is an obvious one. Most people would probably think of using the circle of radius \(r\) that we’re rotating about the \(y\)-axis as the cross-section. It is definitely one of the more obvious choices, however setting up an integral using this is not so easy.

So, what we’ll do is use a cross-section as shown in the sketch to the right above. If we cut the torus perpendicular to the \(y\)-axis we’ll get a cross-section of a ring and finding the area of that shouldn’t be too bad. To do that let’s take a look at the two sketches below.

The sketch to the left is a sketch of the full cross-section. The sketch to the right is more important however. This is a sketch of the circle that we are rotating about the \(y\)-axis. Included is a line representing where the cross-sectional area would be in the torus.

Notice that the inner radius will always be the left portion of the circle and the outer radius will always be the right portion of the circle. Now, we know that the equation of this is,

\[{\left( {x - R} \right)^2} + {y^2} = {r^2}\]

and so if we solve for \(x\) we can get the equations for the left and right sides as shown above in the sketch. This however means that we also now have equations for the inner and outer radii.

\[{\mbox{inner radius}}:x = R - \sqrt {{r^2} - {y^2}} \hspace{0.5in}\hspace{0.25in}{\mbox{outer radius}}:x = R + \sqrt {{r^2} - {y^2}} \]

The cross-sectional area is then,

\[\begin{align*}A\left( y \right) & = \pi {\left( {{\mbox{outer radius}}} \right)^2} - \pi {\left( {{\mbox{inner radius}}} \right)^2}\\ & = \pi \left[ {{{\left( {R + \sqrt {{r^2} - {y^2}} } \right)}^2} - {{\left( {R - \sqrt {{r^2} - {y^2}} } \right)}^2}} \right]\\ & = \pi \left[ {{R^2} + 2R\sqrt {{r^2} - {y^2}} + {r^2} - {y^2} - \left( {{R^2} - 2R\sqrt {{r^2} - {y^2}} + {r^2} - {y^2}} \right)} \right]\\ & = 4\pi R\sqrt {{r^2} - {y^2}} \end{align*}\]

Next, the lowest cross-section will occur at \(y = - r\) and the highest cross-section will occur at \(y = r\) and so the limits for the integral will be \( - r \le y \le r\).

The integral giving the volume is then,

\[V = \int_{{ - r}}^{r}{{4\pi R\sqrt {{r^2} - {y^2}} \,dy}} = 2\int_{0}^{r}{{4\pi R\sqrt {{r^2} - {y^2}} \,dy}} = 8\pi R\int_{0}^{r}{{\sqrt {{r^2} - {y^2}} \,dy}}\]

Note that we used the fact that because the integrand is an even function and we’re integrating over \(\left[ { - r,r} \right]\) we could change the lower limit to zero and double the value of the integral. We saw this fact back in the Computing Definite Integrals section.

We’ve now reached the second really tricky part of this example. With the knowledge that we’ve currently got at this point this integral is not possible to do. It requires something called a trig substitution and that is a topic for Calculus II. Luckily enough for us, and this is the tricky part, in this case we can actually determine the integral’s value using what we know about integrals.

Just for a second let’s think about a different problem. Let’s suppose we wanted to use an integral to determine the area under the portion of the circle of radius \(r\) and centered at the origin that is in the first quadrant. There are a couple of ways to do this, but to match what we’re doing here let’s do the following.

We know that the equation of the circle is \({x^2} + {y^2} = {r^2}\) and if we solve for \(x\) the equation of the circle in the first (and fourth for that matter) quadrant is,

\[x = \sqrt {{r^2} - {y^2}} \]

If we want an integral for the area in the first quadrant we can think of this area as the region between the curve \(x = \sqrt {{r^2} - {y^2}} \) and the \(y\)-axis for \(0 \le y \le r\) and this is,

\[A = \int_{0}^{r}{{\sqrt {{r^2} - {y^2}} \,dy}}\]

In other words, this integral represents one quarter of the area of a circle of radius \(r\) and from basic geometric formulas we now know that this integral must have the value,

\[A = \int_{0}^{r}{{\sqrt {{r^2} - {y^2}} \,dy}} = \frac{1}{4}\pi {r^2}\]

So, putting all this together the volume of the torus is then,

\[V = 8R\pi \int_{0}^{r}{{\sqrt {{r^2} - {y^2}} \,dy}} = 8\pi R\left( {\frac{1}{4}\pi {r^2}} \right) = 2R{\pi ^2}{r^2}\]