(Arc Length and Surface Area) Compute the surface area of revolution of 𝑦=(4−𝑥^(2/3))^(3/2) about the x-axis over the interval [3,8].

Answer 1

We use I[3,8] for the integral from 3 to 8 and E[3,8] for the evaluation from 3 to 8

I[3,8]2π f(x) √(1 + f'²(x)dx dx

f(x) = (4 - x^2/3)^3/2

f'(x) = 3/2 * -2/3x^-1/3(4 - x^2/3)^1/2 = -x^-1/3(4 - x^2/3)^1/2

1 + f'²(x) = 1 + x^-2/3(4 - x^2/3) = 1 + 4x^-2/3 - 1 = 4x^-2/3

Then, √4x^-2/3 = 2x^-1/3

Now, we plug into I[3,8]2π f(x) √(1 + f'²(x)dx dx

We have I[3,8]2π (4 - x^2/3)^3/2 2x^-1/3 dx =

-12/5π (4 - x^2/3)^5/2 E[3,8] =

12/5π (4 - 3^2/3)^5/2