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

MarkFL

Staff member
Here is the question:

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:

I have posted a link there to this topic so the OP can find my response.

MarkFL

Staff member
Hello Henry,

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

$x(t)=3t^2$

$y(t)=2t^3$

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:

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

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

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

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

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

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

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

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

ZaidAlyafey

Well-known member
MHB Math Helper
Just a remark to reduce computations ..

Here we don't need a geometric substitution :

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

rewrite as

$$S=18\pi\int_0^5 2t \cdot t^2\sqrt{1+t^2}\,dt$$

We can use the substitution

$$u=1+t^2 \,\,\Rightarrow \,\, t^2=u-1$$

so we have $$du=2t\, dt$$

The integral becomes as the following :

$$S=18\pi \int_1^{26} (u-1)\sqrt{u}\,du$$

$$S=18\pi \int_1^{26} \sqrt{u^3}-\sqrt{u}\,du$$