Calculate the volume of Torus using cylindrical shells
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Solution
To start, let’s talk a little bit about Torus. We get a Torus by revolving a circle in a 3D space around an axis which is in the same plane as the circle, like it is shown in the next figure
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Now, let’s find out how to calculate the Torus volume. To do that we take the circle of radius \(a\) and center situated at a distance of \(R\) from the axis. We divide the circle into infinitesimal slices, and we consider only one slice of width \(\text{d}x\) and height \(2y\) like it is shown in \(\color{green} {green}\) in the next figure
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The idea is when we revolve the circle around the axis we get the Torus and when we revolve the green slice around the axis we get a cylindrical shell. The next figure explains that
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Now, we need to calculate the volume of the cylindrical shell. The origin of the coordinate system is a point of the axis like it is shown in the figure above. The equation of the circle is: \[\left(x-R\right)^{2}+y^{2}=a^{2}\Rightarrow y^{2}=a^{2}-\left(x-R\right)^{2}\] \[\Rightarrow|y|=\sqrt{a^{2}-\left(x-R\right)^{2}}\] If we want to work on the top part of the circle, then \(y\geq0\) therefore \[y=\sqrt{a^{2}-\left(x-R\right)^{2}}\] Let \(\text{d}s\) be the area of the circle slice, \[\text{d}s=\left(2y\right)\text{d}x\] Now, let’s considerate the next figure that shows a cylindrical shell obtained by revolving the circle slice about the axis by \(2\pi\)
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The circle slice is situated at a distance of \(x\) from the axis. Let \(\text{d}v\) be the volume of the cylindrical shell, thus \[\text{d}v=2\pi x\text{d}s\] \[=2\pi x\left(2y\right)\text{d}x\] \[=4\pi x \sqrt{a^{2}-\left(x-R\right)^{2}}\text{d}x\] For \(R-a\leq x\leq R+a\) \[v=\int_{R-a}^{R+a} 4\pi x \sqrt{a^{2}-\left(x-R\right)^{2}}\text{d}x\] Let’s change the variable of integration, let \[\sin\theta=\frac{x-R}{a}\Rightarrow x-R=a\sin\theta\] \[a^{2}-\left(x-R\right)^{2}=a^{2}\left(1-\left(\sin\theta\right)^{2}\right)=a^{2}\left(\cos\theta\right)^{2}\] When \(x=R-a\Leftrightarrow\theta=-\frac{\pi}{2}\) and when \(x=R+a\Leftrightarrow\theta=\frac{\pi}{2}\) \[\text{d}x=a\cos\theta\text{d}\theta\] And then let’s return to the integral \[v=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}4\pi\left(R+a\sin\theta\right)a\left(\cos\theta\right) a\left(\cos\theta\right)\text{d}\theta\] \[=4\pi a^{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(R\left(\cos\theta\right)^{2}+a\sin\theta\left(\cos\theta\right)^{2}\right)\text{d}\theta\] \[=4\pi a^{2}\left(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}R\left(\cos\theta\right)^{2}\text{d}\theta+a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\theta\left(\cos\theta\right)^{2}\text{d}\theta\right)\] \[=4\pi a^{2}\left(\frac{R}{2}\left(\frac{\sin2\theta}{2}+\theta\right)_\frac{-\pi}{2}^\frac{\pi}{2}-a\left(\frac{\left(\cos\theta\right)^{3}}{3}\right)_\frac{-\pi}{2}^\frac{\pi}{2}\right)\] \[=4\pi a^{2}\left(\frac{R}{2}\left(\frac{\sin\left(2\frac{\pi}{2}\right)}{2}+\frac{\pi}{2}-\frac{\sin\left(2\left(-\frac{\pi}{2}\right)\right)}{2}+\frac{\pi}{2}\right)-a\left(\frac{\left(\cos\frac{\pi}{2}\right)^{3}}{3}-\frac{\left(\cos\left(-\frac{\pi}{2}\right)\right)^{3}}{3}\right)\right)\] \[=4\pi a^{2}\left(\frac{R}{2}\pi\right)\] \[\large \color{olive} {v=2 \pi^{2}a^{2}R}\]
Home -> Solved problems -> Volume of torus using cylindrical shells
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Home -> Solved problems -> Volume of torus using cylindrical shells