Two methods to compute the surface area of torus
by hinn, Feb 5, 2013, 12:19 PM
A torus is obtained by revolving around the y-axis the circle
![\[(x-b)^2 +y^2 =a^2 ~~(0<a\leq b)\]](//latex.artofproblemsolving.com/3/c/d/3cd51b287d2770a119e91da31b473c0f219e7aae.png)
Show that the surface area of the torus.
Here we show two methods to compute it. One is universal, it comes from the reference answer. Inspried by the first method, we can get a more concise one.
Method 1
Firstly, let's compute the area on upper plane, then total area is 2 times of the area on the upper plane.
Secondly, consider the
(suppose the coresponding coordinate on x-axis is x) on the surface, when it revolves around the y-axis, the zonal radius is 
, so the zonal area is 
. And the range of 
is ![$[b-a, b+a]$](//latex.artofproblemsolving.com/4/e/7/4e73e5f60673026ae4e110392dbb79a026ae6e0d.png)
. In this case, the area on the upper plane is
![\[\int_{b-a}^{b+a}2\pi x ds\]](//latex.artofproblemsolving.com/4/e/1/4e18547990c029bb966e10f78177b741b48da941.png)
So the total area of torus is
![\[2\int_{b-a}^{b+a}2\pi x ds\]](//latex.artofproblemsolving.com/f/0/9/f09982f6abbde46e89686ef71b10b74a9522e68a.png)
Recall
and![$\frac{dy}{dx}=\frac{-(x-b)}{[a^2-(x-b)^2]^{1/2}}$](//latex.artofproblemsolving.com/b/6/4/b643191097679bf43ac5e5e1f7a1b34175f3433c.png)
, then $1+\left(\frac{dy}{dx}}\right)^2=\frac{a^2}{a^2-(x-b)^2}$.
So
![\[ds=\frac{a}{\sqrt{a^2-(x-b)^2}}dx\]](//latex.artofproblemsolving.com/3/1/d/31dcc8b50d5961692e6e1f009d0a9f1df6ae15ab.png)
.
Therefore
![\[S=\int_{b-a}^{b+a}2\pi x\frac{a}{\sqrt{a^2-(x-b)^2}}dx=4\pi a\int_{b-a}^{b+a} \frac{x}{\sqrt{a^2-(x-b)^2}}dx\]](//latex.artofproblemsolving.com/4/4/2/4420043adeb59a41afe913f8c19b5e842ec33a92.png)
Let
, then 
,
![\[\sin\theta=\frac{x-b}{a},~~\theta=\arcsin\left(\frac{x-b}{a}\right)\]](//latex.artofproblemsolving.com/0/d/2/0d2ce6f4393fea8ee5d9c75ed2eb64fe95ab685c.png)
Note that when
, we have
and when
,
Consequently
![\begin{align*}
S&=4\pi a\int_{-\pi/2}^{\pi/2}\frac{b+a\sin \theta}{a\cos \theta}a\cos \theta d\theta\\
&=4\pi a\left[b\cdot \frac{\pi}{2}-b\cdot\left(-\frac{\pi}{2}\right)-0+0\right]\\
&=4\pi^2 ab
\end{align*}](//latex.artofproblemsolving.com/8/2/5/8256470c5909e50813012c0f522da46426e9a7f6.png)
Method 2
Consider the cross section, its premeter is
. When it involves y-axis, its surface area is
![\[S=\int_{0}^{2\pi b}2\pi a d\alpha=4\pi^2 ab\]](//latex.artofproblemsolving.com/d/7/6/d760cbf637ea2a033bdd339348aef3e84942f6d6.png)
Obviously, the second method is much more concise than the first one. At the coming tutorial time, I will introduce it briefly.
![\[(x-b)^2 +y^2 =a^2 ~~(0<a\leq b)\]](http://latex.artofproblemsolving.com/3/c/d/3cd51b287d2770a119e91da31b473c0f219e7aae.png)
Show that the surface area of the torus.
Here we show two methods to compute it. One is universal, it comes from the reference answer. Inspried by the first method, we can get a more concise one.
Method 1
Firstly, let's compute the area on upper plane, then total area is 2 times of the area on the upper plane.
Secondly, consider the
(suppose the coresponding coordinate on x-axis is x) on the surface, when it revolves around the y-axis, the zonal radius is 
, so the zonal area is 
. And the range of 
is ![$[b-a, b+a]$](http://latex.artofproblemsolving.com/4/e/7/4e73e5f60673026ae4e110392dbb79a026ae6e0d.png)
. In this case, the area on the upper plane is![\[\int_{b-a}^{b+a}2\pi x ds\]](http://latex.artofproblemsolving.com/4/e/1/4e18547990c029bb966e10f78177b741b48da941.png)
So the total area of torus is
![\[2\int_{b-a}^{b+a}2\pi x ds\]](http://latex.artofproblemsolving.com/f/0/9/f09982f6abbde46e89686ef71b10b74a9522e68a.png)
Recall
\[ds=\sqrt{1+\left(\frac{dy}{dx}}\right)^2} dx\],and
![$\frac{dy}{dx}=\frac{-(x-b)}{[a^2-(x-b)^2]^{1/2}}$](http://latex.artofproblemsolving.com/b/6/4/b643191097679bf43ac5e5e1f7a1b34175f3433c.png)
, then $1+\left(\frac{dy}{dx}}\right)^2=\frac{a^2}{a^2-(x-b)^2}$.So
![\[ds=\frac{a}{\sqrt{a^2-(x-b)^2}}dx\]](http://latex.artofproblemsolving.com/3/1/d/31dcc8b50d5961692e6e1f009d0a9f1df6ae15ab.png)
.Therefore
![\[S=\int_{b-a}^{b+a}2\pi x\frac{a}{\sqrt{a^2-(x-b)^2}}dx=4\pi a\int_{b-a}^{b+a} \frac{x}{\sqrt{a^2-(x-b)^2}}dx\]](http://latex.artofproblemsolving.com/4/4/2/4420043adeb59a41afe913f8c19b5e842ec33a92.png)
Let
, then 
,![\[\sin\theta=\frac{x-b}{a},~~\theta=\arcsin\left(\frac{x-b}{a}\right)\]](http://latex.artofproblemsolving.com/0/d/2/0d2ce6f4393fea8ee5d9c75ed2eb64fe95ab685c.png)
Note that when
, we have\[a\cos \theta=0, \and ~~\theta=\arcsin(1)=\frac{\pi}{2}\],and when
,\[a\cos \theta=0, \and~~ \theta=\arcsin(-1)=-\frac{\pi}{2}\].Consequently
![\begin{align*}
S&=4\pi a\int_{-\pi/2}^{\pi/2}\frac{b+a\sin \theta}{a\cos \theta}a\cos \theta d\theta\\
&=4\pi a\left[b\cdot \frac{\pi}{2}-b\cdot\left(-\frac{\pi}{2}\right)-0+0\right]\\
&=4\pi^2 ab
\end{align*}](http://latex.artofproblemsolving.com/8/2/5/8256470c5909e50813012c0f522da46426e9a7f6.png)
Method 2
Consider the cross section, its premeter is
. When it involves y-axis, its surface area is![\[S=\int_{0}^{2\pi b}2\pi a d\alpha=4\pi^2 ab\]](http://latex.artofproblemsolving.com/d/7/6/d760cbf637ea2a033bdd339348aef3e84942f6d6.png)
Obviously, the second method is much more concise than the first one. At the coming tutorial time, I will introduce it briefly.
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This post has been edited 7 times. Last edited by hinn, Feb 8, 2013, 3:39 PM
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