# Integral - I can't get the right answer

1. Nov 29, 2006

### Sane

Find the volume of the solid obtained when the given region is rotated about the x-axis.

Under $y = (\sin{x})^{\frac{3}{2}}$ between 0 and pi.

The radius is... $r = (\sin{x})^{\frac{3}{2}}$

Then the area for any sample is... $A (x) = \pi((\sin{x})^{\frac{3}{2}})^{2}$

Simplifying to... $A (x) = \pi(\sin{x})^{3}$

Integrate between 0 and pi to get the volume...

$$V = \pi \int_{0}^{\pi} (\sin{x})^{3}$$

$$V = \pi [ \frac{(\sin{x})^{4}}{4\cos{x}} ]_{0}^{\pi}$$

But... sin(pi) and sin(0) both equal 0, making the volume 0. But it's actually (4/3)(pi). What am I missing?

2. Nov 29, 2006

### neutrino

That's not the right way to integrate...

hint: Split (sin(x))^3 as (sin(x))^2.sin(x)

3. Nov 29, 2006

### HallsofIvy

Staff Emeritus
You are missing the fact that
$$\int(\sin{x})^{3}dx$$
is NOT
$$\frac{(\sin{x})^{4}}{4\cos{x}}$$
Since there was not a cosine in the original integrand, you can't simply put one in. In particular,if you try to differentiate
$$\frac{(\sin{x})^{4}}{4\cos{x}}$$
you do not get $sin^3(x)$ because you have to use the quotient rule and differentiate that new "cos(x)" also.

You surely learned how to do integrals of odd powers of sine and cosine long ago: separate out one "sin x" to use with the derivative:
[tex]\int sin^2(x) sin(x)dx= \int (1- cos^2(x)) sin(x)dx[/itex]
and let u= cos(x).

(Blast you, neutrino, you beat me again!)

Last edited: Nov 29, 2006
4. Nov 29, 2006

### AlephZero

I think you got confused doing the integral.

It's true that Integral sin^3x cos x dx = 1/4 sin^4 x

But you can't just "divide by cos x" and say Integral sin^3 x = 1/4 sin^4 x / cos x. Check out the ways you have learned to do integrals like sin^3 x.

5. Nov 29, 2006

### Sane

Ahhh shoot! It didn't even strike me to use substitution. I was being ignorant to the fact that it would require the quotient rule to derivate what I thought was the integral... Thank-you very much! I'm so relieved!

6. Nov 29, 2006

### neutrino

The word is 'differentiate'. :)