Answer
$\frac{22776\pi\sqrt{26}}{5}+\frac{24\pi}{5}$
Work Step by Step
Recall: The area of the surface obtained by rotating the curve $x=f(t)$ and $y=g(t)$ for $a\leq t\leq b$ about the $y-$axis is formulated by the integral $S=\int_a^b2\pi xds$ where $ds=\sqrt{(dx/dt)^2+(dy/dt)^2}dt$
Given: $f(t)=3t^2$ and $g(t)=2t^3$ for $0\leq t\leq 5$
Find $ds$:
$ds=\sqrt{(dx/dt)^2+(dy/dt)^2}dt=\sqrt{(6t)^2+(6t^2)^2}dt=\sqrt{36t^2+36t^4}dt=6t\sqrt{1+t^2}dt$
Evaluate the area of the surface:
$S=\int_0^52\pi(3t^2)\cdot 6t\sqrt{1+t^2}dt$
$S=\int_0^536\pi t^3\sqrt{1+t^2}dt$ (Use the substitution $u=1+t^2\Rightarrow du=2tdt$)
$S=\int_{1+0^2}^{1+5^2}18\pi t^2\sqrt{u}du$ (Use $u=1+t^2\Rightarrow t^2=u-1$)
$S=\int_1^{26}18\pi (u-1)\sqrt{u}du$
$S=\int_1^{26}18\pi u\sqrt{u}-18\pi\sqrt{u}du$
$S=[\frac{36\pi u^2\sqrt{u}}{5}-12\pi u\sqrt{u}]_1^{26}$
$S=(\frac{36\pi 26^2\sqrt{26}}{5}-12\pi 26\sqrt{26})-(\frac{36\pi 1^2\sqrt{1}}{5}-12\pi 1\sqrt{1})$
$S=(\frac{24336\pi\sqrt{26}}{5}-\frac{1560\pi\sqrt{26}}{5})-(\frac{36\pi }{5}-\frac{60\pi}{5})$
$S=\frac{22776\pi\sqrt{26}}{5}+\frac{24\pi}{5}$
Thus, the area of the surface is $\frac{22776\pi\sqrt{26}}{5}+\frac{24\pi}{5}$.
