Question

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

Answer 1

Hint: 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:
\[\int\limits_a^b {\pi \left( {{f^2} - {g^2}} \right)dx} = \int\limits_a^b {\pi \left( {{f^2}\left( x \right)} \right)dx} - \int\limits_a^b {\pi \left( {{g^2}\left( x \right)} \right)dx} \]