What is the expression that gives the area of the rotational surface formed by rotating the curve segment with equations x = t^3 and y = 2t +3 around the y axis, provided that -1<=t<=1?

To find the area of the rotational surface formed by rotating the curve segment around the y-axis, we will use the formula for the surface area of revolution given by:

A = 2\pi\int_{a}^{b} y \sqrt{1 + \left(\frac{dx}{dt}\right)^2} dt

Given that the curve segment is defined by x = t^3 and y = 2t + 3 with a = -1 and b = 1 .

First, find \frac{dx}{dt} :
\frac{dx}{dt} = \frac{d}{dt} (t^3) = 3t^2

Now substitute the values of y , and \frac{dx}{dt} into the formula:
A = 2\pi\int_{-1}^{1} (2t+3) \sqrt{1 + (3t^2)^2} dt
A = 2\pi\int_{-1}^{1} (2t+3) \sqrt{1 + 9t^4} dt

Now integrate the above expression:
A = 2\pi\int_{-1}^{1} (2t+3) \sqrt{1 + 9t^4} dt
A = 2\pi\int_{-1}^{1} (2t\sqrt{1 + 9t^4} + 3\sqrt{1 + 9t^4}) dt