Find the area of the surface obtained by rotating the curve $y=\sqrt[3]{x}$ about $y$-axis for $1 \leq y \leq 4$. Area:

Step 1 ：Given the curve \(y=x^{1/3}\), we are asked to find the area of the surface obtained by rotating this curve about the y-axis for \(1 \leq y \leq 4\).

Step 2 ：The surface area of a solid of revolution is given by the formula: \[A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} dx\]

Step 3 ：For the given function \(f(x) = x^{1/3}\), the derivative is \(f'(x) = \frac{1}{3}x^{-2/3}\).

Step 4 ：Since we are rotating about the y-axis, we need to express everything in terms of y. This gives us \(x = y^3\) and \(dx = 3y^2 dy\).

Step 5 ：Substituting these into the formula gives: \[A = 2\pi \int_{1}^{4} y^3 \sqrt{1 + (3y^2)^2} dy\]

Step 6 ：This is the integral we need to compute.

Step 7 ：Computing the integral gives: \[A = 2\pi \left(-\frac{5\sqrt{10}}{27} + \frac{2305\sqrt{2305}}{54}\right)\]

Step 8 ：Thus, the area of the surface obtained by rotating the curve \(y=x^{1/3}\) about the y-axis for \(1 \leq y \leq 4\) is \(\boxed{-\frac{5\sqrt{10}}{27} + \frac{2305\sqrt{2305}}{54}}\)