# Area Of The Surface Of Revolution

1. Sep 8, 2007

### 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.

Last edited: Sep 8, 2007
2. Sep 8, 2007

### HallsofIvy

Staff Emeritus
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= 9a2 cos4(t)sin2(t)+ 9a2 sin4(t)cos2(t). Factoring out 9a2 cos2(t)sin2(t) leaves cos2(t)+ sin2(t)= 1.
$$\sqrt{x'^2+ y'^2}= \sqrt{9a^2 cos^2(t)sin^2(t)}= 3a cos(t)sin(t)$$
$$y\sqrt{x'^2+ y'^2= 3a^2 cos(t)sin^2(t)$$
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 $\pi/2$.

3. Sep 8, 2007

### 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..

4. Sep 8, 2007

### HallsofIvy

Staff Emeritus
I just type faster!

5. Sep 8, 2007

### Mattofix

Thanks guys, i totally understand it now

6. Sep 8, 2007

### HallsofIvy

Staff Emeritus
I'm glad YOU do!

7. Sep 9, 2007

### Mattofix

and you dont...? how come?

Last edited: Sep 9, 2007
8. Sep 9, 2007

### 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.

Last edited: Sep 9, 2007
9. Sep 9, 2007

### Mattofix

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

10. Sep 9, 2007

### arildno

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

11. Sep 9, 2007

### HallsofIvy

Staff Emeritus
Unfortunately, that happens a lot!