Answer
The surface area of the solid of revolution obtained by rotating the curve $y = x^{(3/2)}/3$, for $0 \leq x \leq 12$ about the y-axis is $\frac{3712\pi}{15}$.
Work Step by Step
The surface area of a solid of revolution obtained by rotating the curve $y = x^{(3/2)}/3$ for $0 \leq x \leq 12$ about the y-axis can be calculated using the formula for the surface area of a solid of revolution:
$$S = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$
In this case, $y = x^{(3/2)}/3$ and $\frac{dy}{dx} = \frac{3}{2} * \frac{x^{(1/2)}}{3}$. Substituting these values into the formula, we get:
$$S = 2\pi \int_{0}^{12} x \sqrt{1 + \left(\frac{3}{2} * \frac{x^{(1/2)}}{3}\right)^2} dx$$
$$= 2\pi \int_{0}^{12} x \sqrt{1 + \left(\frac{x}{4}\right)} dx$$
This integral can be evaluated using substitution. Let $u = 1 + x/4$, $du = dx/4$. The limits of integration change to $1$ and $4$. We get:
$$S = 8\pi \int_{1}^{4} (4u-4) * \sqrt{u} du$$
$$= 8\pi \int_{1}^{4} (4u^{(3/2)} - 4u^{(1/2)}) du$$
$$= 8\pi \left[\frac{8}{5}u^{(5/2)} - \frac{8}{3}u^{(3/2)}\right]_{u=1}^{u=4}$$
$$= 8\pi \left[\frac{8}{5}(32) - \frac{8}{3}(8) - \frac{8}{5}(1) + \frac{8}{3}(1)\right]$$
$$= 8\pi \left[\frac{512}{5} - \frac{64}{3} - \frac{8}{5} + \frac{8}{3}\right]$$
$$= 8\pi \left[\frac{256}{15} - \frac{40}{15} + \frac{40}{15}\right]$$
$$= \frac{3712\pi}{15}.$$
The surface area of the solid of revolution obtained by rotating the curve $y = x^{(3/2)}/3$, for $0 \leq x \leq 12$ about the y-axis is $\frac{3712\pi}{15}$.
