Question

Find the area of the surface generated by revolving the curve about each given axis. $$ x=\frac{1}{3} t^{3}, y=t+1, \quad 1 \leq t \leq 2, \quad y \text { -axis } $$

Short answer

The surface area of the revolution is defined by the definite integral \(2 \pi\int_{1}^{2} \frac{1}{3}(y-1)^3 \sqrt{1+ [(y-1)^2]^2} dy\), which could be solved by using various techniques of integration.

Step-by-step solution

Transform the coordinate equation

Given the equations \(x = \frac{1}{3}t^3\) and \(y = t+1\), we need to write x as a function of y. First, solve the equation \(y = t+1\) for \(t\), obtaining \(t = y-1\). Next, substitute \(t = y-1\) into the first equation and simplify to obtain \(x = \frac{1}{3}(y-1)^3\).

Calculate the derivative of x

To calculate the surface area of revolution, we need the derivative of \(x\) with respect to \(y\). Differentiating \(x = \frac{1}{3}(y-1)^3\) using the chain rule yields \(x' = (y-1)^2\).

Apply the formula for surface area of revolution

Substitute into the surface area of revolution formula, we get \(2 \pi\int_{1}^{2} \frac{1}{3}(y-1)^3 \sqrt{1+ [(y-1)^2]^2} dy\). Use integral calculus to evaluate this expression.

Calculate the definite integral

The computation of definite integral require techniques of integration. This is a non-trivial task and might require numerical approximation.