91Ó°ÊÓ

To find the limits of integration, we need to find the corresponding y-values that correspond to the given x-values. We evaluate the given curve at the bounds \(x = 0\) and \(x = \frac{8}{3}\). \(y(0) = (3 \cdot 0)^{1 / 3} = 0\) \(y \left( \frac{8}{3} \right) = \left( 3 \cdot \frac{8}{3} \right)^{1 / 3} = 2\) So, our limits of integration are \(a = 0\) and \(b = 2\).

Next, we differentiate the given curve with respect to \(x\): \(y = (3x)^{1/3}\) \(\frac{dy}{dx} = \frac{1}{3}(3x)^{-2/3} \cdot 3 = x^{-2/3}\)

Now, we substitute the curve and its derivative into the surface area formula: \(A = 2 \pi \int_{0}^{2} x \sqrt{1 + (x^{-2/3})^2} dy\) To find \(x\) in terms of \(y\), we solve the curve \(y = (3x)^{1/3}\) for x: \(x = \left( \frac{y^3}{3} \right)\) Now, we can rewrite the integral in terms of y: \(A = 2 \pi \int_{0}^{2} \left( \frac{y^3}{3} \right) \sqrt{1 + \frac{y^2}{3}} dy\) And finally, evaluate the definite integral to find the surface area: \(A = 2 \pi \left[\int_{0}^{2} \frac{y^5}{3}dy + \int_{0}^{2} \frac{y^3}{9}dy \right]\) \(A = 2\pi\left[ \frac{1}{18}y^6 + \frac{1}{36}y^4 \right]_{0}^{2}\) \(A = 2\pi\left[\frac{1}{18}(2)^6 + \frac{1}{36}(2)^4 - (0)\right]\) \(A = 2\pi\left[\frac{64}{18} + \frac{16}{36}\right]\) \(A = 2\pi\left[\frac{64}{9} + \frac{8}{9}\right]\) \(A = 2\pi\left[\frac{72}{9}\right]\) \(A = 16\pi\) The area of the surface generated when the curve \(y = (3x)^{1/3}\) is revolved about the y-axis for \(0 \leq x \leq \frac{8}{3}\) is \(16\pi\) square units.