Would anyone care to expound on this problem?

Question

Find the surface area, in square units and correct to the nearest hundredth of a square unit, of the surface of revolution found by revolving the arc segment y = x^2 from (0, 0) to (3, 9), about the x-axis.

The radius would be x².

You're trying to find the circumference of a disk, so it's 2πx².

The slant height is √(1+(2x)²) = √(1+4x²)

∫2πx²√(1+4x²) dx

u = 1+4x²
du = 8x dx
dx = 1/(8x) du
x = √(u-1)/2

So

∫2πx²√(1+4x²) dx
2π∫x²√u 1/(8x)du
π/4∫x√u du
π/4∫√(u-1)/2√u du
π/8∫√(u-1)√u du
π/8∫√(u² - u) du

Then I guess you can do
v = u² - u
dv = 2u du
du = dv / 2u

π/16∫√v/v dv

π/16[2/3v^(3/2)]
π/16[2/3(u² - u)^(3/2)]
π/16[2/3((1+4x²)² - (1+4x²))^(3/2)]

evaluated from x=0 to x=3...

I probably screwed up something.

???

I think the answer is 208.09

a. 3.81
b. 49.42
c. 53.23
d. 208.09
e. 257.51
f. 261.32
g. none of these

Answer 1

You probably should take my attempt at an answer out of it, and just ask it as a question again.

Answer 2

the area of surface of revolution is 2π * integral[a,b] of r(x)*sqrt(1+f'(x)^2) dx.

in this problem, just integrate along y instead of x. so the r(x) becomes r(y) which in your case is sqrt(y), the distance from the axis of revolution to the curve. Should be easy for you from here.