Integral - I can't get the right answer

221
0
Find the volume of the solid obtained when the given region is rotated about the x-axis.

Under [itex]y = (\sin{x})^{\frac{3}{2}}[/itex] between 0 and pi.


The radius is... [itex]r = (\sin{x})^{\frac{3}{2}}[/itex]

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

Simplifying to... [itex]A (x) = \pi(\sin{x})^{3}[/itex]

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

[tex]V = \pi \int_{0}^{\pi} (\sin{x})^{3}[/tex]

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

But... sin(pi) and sin(0) both equal 0, making the volume 0. But it's actually (4/3)(pi). What am I missing?
 
2,002
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Integrate between 0 and pi to get the volume...

[tex]V = \pi \int_{0}^{\pi} (\sin{x})^{3}[/tex]

[tex]V = \pi [ \frac{(\sin{x})^{4}}{4\cos{x}} ]_{0}^{\pi}[/tex]
That's not the right way to integrate...

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

HallsofIvy

Science Advisor
41,626
821
You are missing the fact that
[tex]\int(\sin{x})^{3}dx[/tex]
is NOT
[tex] \frac{(\sin{x})^{4}}{4\cos{x}}[/tex]
Since there was not a cosine in the original integrand, you can't simply put one in. In particular,if you try to differentiate
[tex]\frac{(\sin{x})^{4}}{4\cos{x}} [/tex]
you do not get [itex]sin^3(x)[/itex] 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 by a moderator:

AlephZero

Science Advisor
Homework Helper
6,920
290
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.
 
221
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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! :smile:
 
2,002
2
it would require the quotient rule to derivate what I thought was the integral
The word is 'differentiate'. :)
 

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