Volume integral with two intersecting shapes

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The question I am dealing with has to do with the volume contained by two intersecting shapes I have created this integral and can't find a reasonable way of solving it. What is the best approach to solve this:


[tex]\int_{r=0}^{\pi/2}\int_{u=0}^{2\cos\theta} \sqrt{9-r^2} * r drdu= [/tex]
 

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tiny-tim
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The question I am dealing with has to do with the volume contained by two intersecting shapes I have created this integral and can't find a reasonable way of solving it. What is the best approach to solve this:

Hi bobsmiters! :smile:

[tex]\int_{r=0}^{\pi/2}\int_{u=0}^{2\cos\theta} \sqrt{9-r^2} * r drdu= [/tex]
Your integral doesn't look right. You have u r and theta, but only two ∫s. :confused:

Do you mean
[tex]\int_{\theta=0}^{\pi/2}\int_{u=0}^{2\cos\theta} [/tex] ?

If so, where does r come into it?

I hope r is right, because
[tex]\int \sqrt{9-r^2} * r dr[/tex] is very easy … :smile:
 
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The question I am dealing with has to do with the volume contained by two intersecting shapes I have created this integral and can't find a reasonable way of solving it. What is the best approach to solve this:


[tex]\int_{\theta=0}^{\pi/2}\int_{r=0}^{2\cos\theta} \sqrt{9-r^2} * r drd\theta= [/tex]
There wasn't supposed to be a 'u' in the question
 
tiny-tim
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… one integral at a time …

There wasn't supposed to be a 'u' in the question
ah! … that makes more sense! :smile:

right … when we do double integrals, we always do one integral at a time.

In this case, do the r integral first.

What is ∫r√(9-r^2)dr?

It's actually quite easy if t you think about it. :smile:
 
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[tex]\int_{\theta=0}^{\pi}\int_{r=0}^{2\cos\theta} \sqrt{9-r^2} * r drd\theta= [/tex]

The first integral didn't cause me much of a problem, however when I insert the 2cos(theta) and the '0' I can't figure out how to work through the second integral.

I ended up with:

[tex]\int_{\theta=0}^{\pi}\sqrt{(9-4\cos^2\theta)^3} / (-3) +9 d\theta= [/tex]
 
tiny-tim
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Hi bobsmiters! :smile:

hmm … don't like the look of that …

I'll have a think about it later, but before I do …

You got this ∫∫drdtheta yourself, didn't you?

I was a little surprised when I first saw it because usually when you convert to polar coordinates, you get an extra sintheta for free (because dxdy = sintheta dr dtheta).
(and then I got confused by the du, and forgot about it :confused:)

And that extra sintheta tends to make things much easier.

Are you sure there shouldn't be a sintheta?
 
Vid
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You're thinking of spherical coordinates.

dA = dxdy = rdrdtheta
dV = dxdydz = r^2sin(theta)dr*dtheta*dphi
 
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The original question read: Find the volume of the solid that lies under x^2 + y^2 + z^2 = 9, above the xy-plane and inside the cylinder x^2 + y^2 = 2x.

It is possible that I made a mistake somewhere in getting to that integral.
 
tiny-tim
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You're thinking of spherical coordinates.

dA = dxdy = rdrdtheta
dV = dxdydz = r^2sin(theta)dr*dtheta*dphi
Indeed I am! :smile:
The original question read: Find the volume of the solid that lies under x^2 + y^2 + z^2 = 9, above the xy-plane and inside the cylinder x^2 + y^2 = 2x.

It is possible that I made a mistake somewhere in getting to that integral.
Hi bobsmiters! :smile:

I'm confused … how did you get that … what is your r? :confused:
 

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