Find the area of the surface obtained by rotating the curve?

Question

Find the area of the surface obtained by rotating the curve

y=3x^3

from x=0 to x=7 about the x-axis

Answer 1

This is much easier to write than type...

Start with the formula for surface area of an arc of revolution:

SA = 2π ∫ { r(x) * √(1+[ƒ'(x)]²) dx }

Limits of integration (∫):

x=0 to x=7

r(x) = x³  (distance of curve from axis of rotation)

ƒ(x) = x³  (given function)

ƒ'(x) = 3x²

[ƒ'(x)]² = 9x^4

√(1+[ƒ'(x)]²) } = √(1+ 9x^4)

SA = 2π ∫ { x³ * √(1+ 9x^4) dx }  [∫ x=0 to x=7]

-------------------- -------------------- --------------------

u substitution:

u = 1+ 9x^4

du = 36x³ dx  -->  (du/36) = x³ dx

new limits for u:

@ x = 0, u = 1+ 9(0)^4 = 1

@ x = 7, u = 1+ 9(7)^4 = 21,610  (yes, 21,610!)

SA = 2π ∫ { (du/36) * √(u) }  [∫ u=1 to u=21,610]

SA = (π/18) ∫ { u^(1/2) du }  [∫ u=1 to u=21,610]

-------------------- -------------------- --------------------

Integrate:

SA = (π/18) * [(2/3) * u^(3/2)]  [u=1 to u=21,610]

Evaluate over the limits of u to find surface area:

SA = (π/27) * [21,610^(3/2) - 1^(3/2)]

SA = ...

The answer will be on the order of 10^5.