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

1. Apr 14, 2012

### ross1219

1. The problem statement, all variables and given/known data
Find the area of the surface generated by revolving the curve

2. Relevant equations
x = 5(cos3t), y = 5(sin3t), 0 ≤ t ≤ $\pi$, about the y axis.

3. 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$\pi$$\int$(from 0 to $\pi$/2) 5(cos3t) sqrt[225(cos4t)(sin2t) + 225(sin4t)(cos2t)] dt

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

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

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

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

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

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

S = -150$\pi$ (0 - 1/5)

S = 30$\pi$ units2

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

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

Thank you much.

Last edited: Apr 14, 2012
2. Apr 25, 2012

### 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.