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:
$\displaystyle S=36\pi\int_{\frac{1}{\sqrt{26}}}^{1} \frac{1-u^2}{u^6}\,du$
$\displaystyle S=36\pi\int_{\frac{1}{\sqrt{26}}}^{1} u^{-6}-u^{-4}\,du$
$\displaystyle S=36\pi\left[\frac{u^{-5}}{-5}-\frac{u^{-3}}{-3} \right]_{\frac{1}{\sqrt{26}}}^{1}$
$\displaystyle S=\frac{12\pi}{5}\left[5u^{-3}-3u^{-5} \right]_{\frac{1}{\sqrt{26}}}^{1}$
$\displaystyle S=\frac{12\pi}{5}\left(\left(5(1)^{-3}-3(1)^{-5} \right)- \left(5\left(\frac{1}{\sqrt{26}} \right)^{-3}-3\left(\frac{1}{\sqrt{26}} \right)^{-5} \right) \right)$
$\displaystyle S=\frac{12\pi}{5}\left(5-3-130\sqrt{26}+2028\sqrt{26} \right)$
$\displaystyle S=\frac{12\pi}{5}\left(2+1898\sqrt{26} \right)$
$\displaystyle S=\frac{24\pi}{5}\left(1+949\sqrt{26} \right)$