Question

Find the area of the surface generated by revolving x=t+√{6}, y=t^{2}}{2}+√{6} t+1,√{6} ≤ t ≤ √{6} \) about the ( y )-axis.

Answer 1

To find the area of the surface generated by revolving the given curve about the y-axis, we can use the formula for the surface area of revolution.

Step-by-step explanation:

To find the area of the surface generated by revolving the given curve about the y-axis, we can use the formula for the surface area of revolution:

S = 2π ∫(y * sqrt(1 + (dy/dx)^2)) dx

In this case, the curve is defined by the equations:

x = t + sqrt(6)

y = (t^2)/2 + sqrt(6)*t + 1

First, we need to find dy/dx:

dy/dx = (dy/dt) / (dx/dt)

dy/dx = ((2t + sqrt(6)) / 2) / 1

dy/dx = t + sqrt(6)/2

Substituting this into the surface area formula and integrating over the range sqrt(6) ≤ t ≤ sqrt(6), we can find the area of the surface of revolution.