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Henry's question at Yahoo! Answers concerning a surface of revolution

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Staff member
Feb 24, 2012
Here is the question:

Calculus 2 help me please?

Find the surface area generated by rotating the given curve about the y-axis.

x = 3t^2, y = 2t^3, 0 ≤ t ≤ 5
Here is a link to the question:

Calculus 2 help me please? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
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Feb 24, 2012
Hello Henry,

We are asked to find the surface of rotation of the curve described parametrically by:



with $t$ in $[0,5]$.

Since the axis of rotation is the $y$-axis and $x(t)$ is non-negative on the given interval for $t$, we may use:

$\displaystyle S=2\pi\int_0^5 x(t)\sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dy}{dt} \right)^2}\,dt$

So, we compute:

$\displaystyle \frac{dx}{dt}=6t$

$\displaystyle \frac{dy}{dt}=6t^2$

and we have:

$\displaystyle S=2\pi\int_0^5 3t^2\sqrt{\left(6t \right)^2+\left(6t^2 \right)^2}\,dt$

$\displaystyle S=36\pi\int_0^5 t^3\sqrt{1+t^2}\,dt$

Now, let's use the substitution:

$\displaystyle t=\tan(\theta)\,\therefore\,dt=\sec^2( \theta)\,d \theta$ and we have:

$\displaystyle S=36\pi\int_0^{\tan^{-1}(5)} \tan^3(\theta)\sqrt{1+\tan^2(\theta)}\,sec^2( \theta)\,d \theta$

$\displaystyle S=36\pi\int_0^{\tan^{-1}(5)} \tan^3(\theta)\sec^3(\theta)\,d\theta$

$\displaystyle S=36\pi\int_0^{\tan^{-1}(5)} \frac{\sin(\theta)(1-\cos^2(\theta))}{\cos^6(\theta)}\,d\theta$

Now, let's try the substitution:

$u=\cos(\theta)\,\therefore\,du=-\sin(\theta)\,d \theta)$ and we have:

$\displaystyle S=36\pi\int_{\frac{1}{\sqrt{26}}}^{1} \frac{1-u^2}{u^6}\,du$

$\displaystyle S=36\pi\int_{\frac{1}{\sqrt{26}}}^{1} u^{-6}-u^{-4}\,du$

$\displaystyle S=36\pi\left[\frac{u^{-5}}{-5}-\frac{u^{-3}}{-3} \right]_{\frac{1}{\sqrt{26}}}^{1}$

$\displaystyle S=\frac{12\pi}{5}\left[5u^{-3}-3u^{-5} \right]_{\frac{1}{\sqrt{26}}}^{1}$

$\displaystyle S=\frac{12\pi}{5}\left(\left(5(1)^{-3}-3(1)^{-5} \right)- \left(5\left(\frac{1}{\sqrt{26}} \right)^{-3}-3\left(\frac{1}{\sqrt{26}} \right)^{-5} \right) \right)$

$\displaystyle S=\frac{12\pi}{5}\left(5-3-130\sqrt{26}+2028\sqrt{26} \right)$

$\displaystyle S=\frac{12\pi}{5}\left(2+1898\sqrt{26} \right)$

$\displaystyle S=\frac{24\pi}{5}\left(1+949\sqrt{26} \right)$


Well-known member
MHB Math Helper
Jan 17, 2013
Just a remark to reduce computations ..

Here we don't need a geometric substitution :

[tex]S=36\pi\int_0^5 t^3\sqrt{1+t^2}\,dt[/tex]

rewrite as

[tex]S=18\pi\int_0^5 2t \cdot t^2\sqrt{1+t^2}\,dt[/tex]

We can use the substitution

[tex]u=1+t^2 \,\,\Rightarrow \,\, t^2=u-1 [/tex]

so we have [tex]du=2t\, dt[/tex]

The integral becomes as the following :

[tex]S=18\pi \int_1^{26} (u-1)\sqrt{u}\,du[/tex]

[tex]S=18\pi \int_1^{26} \sqrt{u^3}-\sqrt{u}\,du[/tex]