Answer
$23.450$
Work Step by Step
$x=\frac{1}{3}t^3$, $y=t+1$, $1\leq t \leq 2$, y-axis
Use the area for the surface area of a parametric curve:
$S=2\pi\int_{a}^{b}x(t)\sqrt ((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)dt$
$\frac{dx}{dθ}=t^2$
$\frac{dy}{dθ}=1$
$S=2\pi\int_{1}^{2}(\frac{1}{3}t^3)\sqrt( (t^2)^2+(1)^2)dt$
$S=\frac{2}{3}\pi\int_{1}^{2}(t^3)\sqrt(t^4+1)dt$
Use u-sub:
$let$ $u=t^4+1$
$du=4t^3dt$
$\frac{1}{4}du=t^3dt$
$u(1)=2$
$u(2)=17$
$S=\frac{1}{6}\pi\int_{2}^{17}\sqrt(u)du$
$S=\frac{1}{6}\pi[\frac{2}{3}u^{\frac{3}{2}}]_{2}^{17}$
$S=\frac{1}{9}\pi[u^{\frac{3}{2}}]_{2}^{17}$
$S=\frac{1}{9}\pi[17^{\frac{3}{2}}-2^{\frac{3}{2}}]$
$S\approx23.450$
