Evaluating the integral $S=\int_0^12\pi t^4\sqrt{9t^2+4}dt$

Question

I want to find the surface area $S$ by rotating the curve about the $x$-axis $$ \begin{cases} x = t^3 \\ y = t^2 \end{cases} ,t\in [0,1] $$ At some point I find $$S=\int_0^12\pi t^4\sqrt{9t^2+4}dt$$ and I can't go further to evaluate this integral. Thank you for your help !

Answer 1

Your integral formula for $S$ is incorret. It should be $$\begin{align}S &= \int_0^1 2\pi t^2 \sqrt{9t^4 + 4t^2} dt = \int_0^1 2\pi t^2 \cdot t \sqrt{9t^2 + 4} dt \\&= 2\pi\int_0^1 \color{red}{t^3} \sqrt{9t^2 + 4} dt\end{align}$$

Now simply take $u = 9t^2 + 4$.

Answer 2

HINT:

$$\text{S}=\int_{0}^{1}2\pi t^4\sqrt{9t^2+4}\space\text{d}t=2\pi\int_{0}^{1}t^4\sqrt{9t^2+4}\space\text{d}t=$$

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Substitute $t=\frac{2\tan(u)}{3}$ and $\text{d}t=\frac{2\sec^2(u)}{3}\space\text{d}u$.

Then $\sqrt{9t^2+4}=\sqrt{4\tan^2(u)+4}=2\sec(u)$ and $u=\arctan\left(\frac{3t}{2}\right)$.

This gives a new lower bound $u=\arctan\left(\frac{3\cdot0}{2}\right)=0$ and upper bound $u=\arctan\left(\frac{3}{2}\right)$:

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$$\frac{128\pi}{243}\int_{0}^{\arctan\left(\frac{3}{2}\right)}\tan^4(u)\sec^4(u)\space\text{d}u=\frac{128\pi}{243}\int_{0}^{\arctan\left(\frac{3}{2}\right)}\left[\sec^7(u)-2\sec^5(u)+\sec^3(u)\right]\space\text{d}u$$ $$\frac{128\pi}{243}\left[\int_{0}^{\arctan\left(\frac{3}{2}\right)}\sec^7(u)\space\text{d}u-2\int_{0}^{\arctan\left(\frac{3}{2}\right)}\sec^5(u)\space\text{d}u+\int_{0}^{\arctan\left(\frac{3}{2}\right)}\sec^3(u)\space\text{d}u\right]$$

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After this:

• start using the reduction formula: $$\int\sec^m(u)\space\text{d}u=\frac{\sin(u)\sec^{m-1}(u)}{m-1}+\frac{m-2}{m-1}\int\sec^{m-2}(u)\space\text{d}u$$
• and notice: $$\int\sec(u)\space\text{d}u=\ln\left|\tan(u)+\sec(u)\right|+\text{C}$$