Answer
$$S = \frac {24 \pi} 5( {949} \sqrt {26} + 1 )$$
Work Step by Step
1. Write the formula for the area of the surface obtained by rotating a curve about the $y$-axis.
$$S = \int_{\alpha}^{\beta} 2\pi x \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Calculate: $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = \frac{d(3t^2)}{dt} = 6t$
$\frac{dy}{dt} = \frac{d(2t^3)}{dt} = 6t^2 $
3. Substitute the given values into the formula:
$S = \int_{0}^{5} 2\pi (3t^2) \sqrt {(6t)^2 + (6t^2)^2} \space dt$
$S = \int_{0}^{5} 2\pi (3t^2) \sqrt {36t^2 + 36t^4} \space dt$
$S = \int_{0}^{5} 2\pi (3t^2) \sqrt {36t^2( 1 + t^2)} \space dt$
$S = \int_{0}^{5} 2\pi (18t^3) \sqrt { 1 + t^2} \space dt$
$S = 18 \pi \int_{0}^{5} (2t)(t^2) \sqrt { 1 + t^2} \space dt$
** u = $t^2 + 1 \longrightarrow \frac{du}{dt} = 2t$
** $t^2 = u -1$
** $u_1 = 0^2 + 1 = 1$
** $u_2 = 5^2 + 1 = 26$
$S = 18 \pi \int_{1}^{26} (\frac{du}{dt})(u - 1) \sqrt { u} \space dt$
$S = 18 \pi \int_{1}^{26} (u\sqrt u - \sqrt u) \space du$
$S = 18 \pi \int_{1}^{26} (u^{3/2} - u^{1/2}) \space du$
$S = 18 \pi [\frac 2 5 u^{5/2} - \frac 2 3 u^{3/2}]_1^{26}$
$S = 18 \pi [\frac 2 5 (26)^{5/2} - \frac 2 3 (26)^{3/2}] - 18 \pi [[\frac 2 5 (1)^{5/2} - \frac 2 3 (1)^{3/2}]]$
$S = 18 \pi [\frac {3796} {15} \sqrt {26}] - 18 \pi [\frac {-4} {15} ] $
$S = \frac {63328 \pi} {15} \sqrt {26} + \frac {72 \pi} {15} $
$S = 72( \frac {949 \pi} {15} \sqrt {26} + \frac {\pi} {15}) $
$S = \frac {24 \pi} 5( {949} \sqrt {26} + 1) $
