Area of Surface of Revolution of Plane Curve - Use 1/2 Interval?

[ross1219]

Homework Statement

Find the area of the surface generated by revolving the curve

Homework Equations

x = 5(cos³t), y = 5(sin³t), 0 ≤ t ≤ [itex]\pi[/itex], about the y axis.

The Attempt at a Solution

x' = -15(cos²t)(sin t)
y' = 15(sin²t)(cos t)

(I think this forms the top part of an astroid)

[x'(t)]² = 225(cos⁴t)(sin²t)
[y'(t)]² = 225(sin⁴t)(cos²t)

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos³t) sqrt[225(cos⁴t)(sin²t) + 225(sin⁴t)(cos²t)] dt

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos³t) sqrt[225(cos²t)(sin²t)*(cos²t + sin²t)] dt

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos³t) sqrt[225(cos²t)(sin²t)(1)] dt

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos³t)*15(cos t)(sin t) dt

S = 150[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) (cos⁴t)(sin t) dt

S = -150[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) (cos⁴t)(-sin t) dt

S = -150[itex]\pi[/itex] [(cos⁵t) / 5] from 0 to [itex]\pi[/itex]/2

S = -150[itex]\pi[/itex] (0 - 1/5)

S = 30[itex]\pi[/itex] units²

Please excuse my formatting. My basic question is, was I correct to take the integral only from 0 to [itex]\pi[/itex]/2, even though the interval given for the plane curve is from 0 to [itex]\pi[/itex]?

My secondary question is, do you see anything else I messed up?

Thank you much.

 

[ross1219]

Follow-up: The instructor admitted that this question was not valid, because the curve does not lie entirely on one side of the axis of revolution.