(adsbygoogle = window.adsbygoogle || []).push({}); 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(cos^{3}t), y = 5(sin^{3}t), 0 ≤ t ≤ [itex]\pi[/itex], about the y axis.

3. The attempt at a solution

x' = -15(cos^{2}t)(sin t)

y' = 15(sin^{2}t)(cos t)

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

[x'(t)]^{2}= 225(cos^{4}t)(sin^{2}t)

[y'(t)]^{2}= 225(sin^{4}t)(cos^{2}t)

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos^{3}t) sqrt[225(cos^{4}t)(sin^{2}t) + 225(sin^{4}t)(cos^{2}t)] dt

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos^{3}t) sqrt[225(cos^{2}t)(sin^{2}t)*(cos^{2}t + sin^{2}t)] dt

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos^{3}t) sqrt[225(cos^{2}t)(sin^{2}t)(1)] dt

S = 2[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) 5(cos^{3}t)*15(cos t)(sin t) dt

S = 150[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) (cos^{4}t)(sin t) dt

S = -150[itex]\pi[/itex][itex]\int[/itex](from 0 to [itex]\pi[/itex]/2) (cos^{4}t)(-sin t) dt

S = -150[itex]\pi[/itex] [(cos^{5}t) / 5] from 0 to [itex]\pi[/itex]/2

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

S = 30[itex]\pi[/itex] units^{2}

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.

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

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