Question

How do you use an integral to find the volume of a solid torus?

Answer 1

Here we will use the fact that a torus is obtained by rotating a circular region about the $x$-axis. We will use the washer method to form an expression of obtaining a torus. Finally, we will simplify the expression to get the required answer.

Complete step-by-step answer:
The volume of a torus whose radius of circular cross-section is $r$ and the radius of the circle traced by the center of the cross-section is $R$ is given as, $V = 2{\pi ^2}{r^2}R$.
So we have to use integral to get the above value of the volume of a solid torus.
As we know a torus is obtained by rotating the circular region about the x-axis so it can be written as,
${x^2} + {\left( {y - R} \right)^2} = {r^2}$
Subtracting ${x^2}$ from both the sides, we get
$\begin{array}{l} \Rightarrow {\left( {y - R} \right)^2} = {r^2} - {x^2}\\\ \Rightarrow \left( {y - R} \right) = \pm \sqrt {{r^2} - {x^2}} \end{array}$
Adding $R$ on both the sides, we get
$\Rightarrow y = R \pm \sqrt {{r^2} - {x^2}}$
So, we have two regions as,
$y = R + \sqrt {{r^2} - {x^2}}$ and $y = R - \sqrt {{r^2} - {x^2}}$
As we have two regions in-between which the circular Region is present, so here we can use Washer Method.
According to Washer Method, the volume of the solid of revolution can be expressed as $\int\limits_a^b {\pi \left( {{f^2} - {g^2}} \right)dx} = \int\limits_a^b {\pi \left( {{f^2}\left( x \right)} \right)dx} - \int\limits_a^b {\pi \left( {{g^2}\left( x \right)} \right)dx}$
Here, $y = f\left( x \right)$and$y = g\left( x \right)$ are two curves.
In this case we have $f\left( x \right) = R + \sqrt {{r^2} - {x^2}}$ and $g\left( x \right) = R - \sqrt {{r^2} - {x^2}}$.
Substituting the above values in the formula $\int\limits_a^b {\pi \left( {{f^2} - {g^2}} \right)dx} = \int\limits_a^b {\pi \left( {{f^2}\left( x \right)} \right)dx} - \int\limits_a^b {\pi \left( {{g^2}\left( x \right)} \right)dx}$ we get,
$\int\limits_a^b {\pi \left( {{f^2} - {g^2}} \right)dx} = \int\limits_a^b {\pi {{\left( {R + \sqrt {{r^2} - {x^2}} } \right)}^2}dx} - \int\limits_a^b {\pi {{\left( {R - \sqrt {{r^2} - {x^2}} } \right)}^2}dx}$
Substituting the limit of $x$ from $r$ to $- r$ and $\int\limits_{ - r}^r {\pi \left( {{f^2} - {g^2}} \right)dx} = V$, we get
$\Rightarrow V = \int\limits_{ - r}^r {\pi {{\left( {R + \sqrt {{r^2} - {x^2}} } \right)}^2}dx} - \int\limits_{ - r}^r {\pi {{\left( {R - \sqrt {{r^2} - {x^2}} } \right)}^2}dx}$
Simplifying the equation, we get
$\Rightarrow V = \pi \int\limits_{ - r}^r {\left( {\left( {{{\left( {R + \sqrt {{r^2} - {x^2}} } \right)}^2} - {{\left( {R - \sqrt {{r^2} - {x^2}} } \right)}^2}} \right)} \right)} dx$
Using the algebraic identities, we get
$\Rightarrow V = \pi \int\limits_{ - r}^r {\left( {{R^2} + \left( {{r^2} - {x^2}} \right) + 2 \times R\left( {{r^2} - {x^2}} \right) - \left( {{R^2} - 2 \times R \times \left( {{r^2} - {x^2}} \right) + \left( {{r^2} - {x^2}} \right)} \right)} \right)} dx$
Multiplying the terms, we get
$\Rightarrow V = \pi \int\limits_{ - r}^r {\left( {{R^2} + 2R\left( {{r^2} - {x^2}} \right) + \left( {{r^2} - {x^2}} \right) - {R^2} + 2R\left( {{r^2} - {x^2}} \right) - \left( {{r^2} - {x^2}} \right)} \right)dx}$
Adding and subtracting the like terms, we get
$\Rightarrow V = 4\pi R\int\limits_{ - r}^r {\left( {{r^2} - {x^2}} \right)dx}$
As we can see that above integral is equivalent to the area of a semicircle having radius as $r$ we get,
$\begin{array}{l} \Rightarrow V = 4\pi R \times \dfrac{1}{2}\pi {r^2}\\\ \Rightarrow V = 2{\pi ^2}{r^2}R\end{array}$
As we can see the value we got above is same as the volume of a torus $V = 2{\pi ^2}{r^2}R$, therefore the Volume of solid torus is $2{\pi ^2}{r^2}R$.

Note:
A solid torus is a shape formed when we sweep a disk around a circle. We can visualize a solid torus as a toroid embedded in 3-space.
According to the Washer Method: Let us consider a region $R$ between two curves $y = f\left( x \right)$ and $y = g\left( x \right)$ where both the curves are non-negative and $g\left( x \right) \le f\left( x \right)$ for interval from x = a$$$$x = b. So, the volume traced when $R$ is rotated about the x-axis is given as:
$\int\limits_a^b {\pi \left( {{f^2} - {g^2}} \right)dx} = \int\limits_a^b {\pi \left( {{f^2}\left( x \right)} \right)dx} - \int\limits_a^b {\pi \left( {{g^2}\left( x \right)} \right)dx}$