Area Of The Surface Of Revolution

[Mattofix]

I have tried this question (http://img524.imageshack.us/img524/9539/scan0001pa1.png) a number of times and always use the formula

S = 2*pi*int( y*sqrt( (dx/dt)^2+(dy/dt)^2 ) dt

i always get S=6*pi*a^2[1/5*(sin t)^5] 0<t<pi, and because sin 0 = 0 and sin pi = 0 the answer i get is 0. If you were to replace the y in the equation with x then i get the correct answer but that would not be the formula for rotation about the x-axis.

Any help would be much appreciated.

 

[HallsofIvy]

x= a cos³(t), y= a sin³(t) so x'= 3a cos²(t)sin(t) and y'= a sin²(t)cos(t). x'²+ y'²= 9a<sup>2 cos4(t)sin2(t)+ 9a2 sin4(t)cos2(t). Factoring out 9a2 cos2(t)sin2(t) leaves cos2(t)+ sin2(t)= 1.
\sqrt{x&#039;^2+ y&#039;^2}= \sqrt{9a^2 cos^2(t)sin^2(t)}= 3a cos(t)sin(t)
y\sqrt{x&#039;^2+ y&#039;^2= 3a^2 cos(t)sin^2(t)
After substitution, I get
6\pi a^2\int_{-\pi}^{pi}(sin^2(t)cos(t)) dt

You are right- the substitution u= sin(t) gives u= 0 at both points.

Aha! I think I see the problem. cos(t), for t> \pi/2, is negative. The square root of cos2(t) is NOT cos(t) for t> \pi/2! More generally, \sqrt{x^2}= |x|, not x.

Use symmetry. integrate from 0 to \pi/2. That will give you a positive value. Because the integral from 0 to \pi is 0, the integral from \pi/2 to pi must be the negative of that (you can, if you wish, integrate and see). The true value of the integral from 0 to \pi, using absolute value, must be twice the integral from 0 to \pi/2. And, of course, the actual surface area, the integral from -\pi to \pi, is 4 times the integral from 0 to \pi/2.</sup>

 

[arildno]

Hmm..we should use absolute value signs here!

Thereby, we get the integral:
6a^{2}\pi\int_{0}^{\pi}\sin^{4}(t)|\cos(t)|dt
where I have utilized that sin(t) is positive on the entire interval.

EDIT:
Okay, Halls found the flaw as well..

 

[HallsofIvy]

I just type faster!

 

[Mattofix]

Thanks guys, i totally understand it now

 

[HallsofIvy]

I'm glad YOU do!

 

[Mattofix]

and you dont...? how come?

 

[arildno]

I think Halls is flabbergasted as to why he forgot the absolute value sign in the first place!

Besides, in his expression, he mistakenly used y=asin(t), rather than the correct third power of sine in the y-expression.

 

[Mattofix]

Besides, in his expression, he mistakenly used y=asin(t), rather than the correct third power of sine in the y-expression.

... and that would keep on giving him the wrong answer therefore not understanding it?

 

[arildno]

I think he was merely angered at himself for making a dumb mistake, that's all.

 

[HallsofIvy]

Unfortunately, that happens a lot!