Questions: Find the area of the surface generated by revolving the parametric curve x = 2t - 2, y = t^2 - 1 about the x-axis. 2π ∫ y √(dx/dt)^2 + (dy/dt)^2 dt

Solution Steps

To find the area of the surface generated by revolving the parametric curve \( x = 2t - 2 \) and \( y = t^2 - 1 \) about the x-axis, we need to use the formula for the surface area of a parametric curve revolved around the x-axis. The formula is:

\[ 2\pi \int y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

1. Compute the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
2. Substitute \( y \), \(\frac{dx}{dt}\), and \(\frac{dy}{dt}\) into the formula.
3. Evaluate the integral over the given range of \( t \).

The parametric equations for the curve are given by: \[ x = 2t - 2 \] \[ y = t^2 - 1 \]

We compute the derivatives of \( x \) and \( y \) with respect to \( t \): \[ \frac{dx}{dt} = 2 \] \[ \frac{dy}{dt} = 2t \]

The integrand for the surface area formula is: \[ \text{integrand} = y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = (t^2 - 1) \sqrt{4 + 4t^2} \]

The surface area \( S \) is given by: \[ S = 2\pi \int_{t_1}^{t_2} (t^2 - 1) \sqrt{4 + 4t^2} \, dt \] After evaluating the integral over the range \( t_1 = 0 \) to \( t_2 = 1 \), we find: \[ S = 2\pi \left(-\frac{5 \log(1 + \sqrt{2})}{4} - \frac{\sqrt{2}}{4}\right) \]

Final Answer