Question

Consider the curve below.

x=t, y =t⁴, 0 ≤ t ≤ 1

Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the x-axis.
1
∫ ___________ dt
0

Answer 1

The integral for the surface area of the curve rotated about the x-axis is set up using the formula for the surface area of a revolution with given parametric equations, resulting in ∫_{0}^{1} 2πt⁴ √(1 + (4t³)²) dt.

Step-by-step explanation:

The task is to set up an integral for the surface area of a curve rotated about the x-axis. The given parametric equations are x = t and y = t⁴, where t ranges from 0 to 1. To find the surface area generated by rotating a curve about the x-axis, we can use the formula for the surface area of revolution, which involves an integral of the form:

∫ 2πy ∙ √(1 + (dy/dx)²) dx

First, we find the derivative dy/dx by differentiating y = t⁴ with respect to t and then using x = t to express dy/dx solely in terms of x. This leads to the derivative dy/dt = 4t³. Since dx/dt = 1, dy/dx = (dy/dt) / (dx/dt) = 4t³. Next, we calculate √(1 + (dy/dx)²), which becomes √(1 + (4t³)²). The integral for the surface area is then:

∫_{0}^{1} 2πt⁴ √(1 + (4t³)²) dt

Remember to not evaluate this integral, as the question has specifically asked only to set up the integral for the surface area.