Volume integral with two intersecting shapes

In summary, the conversation is about finding the volume of a solid contained by two intersecting shapes. The integral \int_{r=0}^{\pi/2}\int_{u=0}^{2\cos\theta} \sqrt{9-r^2} * r drdu is incorrect and the correct integral is \int_{\theta=0}^{\pi/2}\int_{r=0}^{2\cos\theta} \sqrt{9-r^2} * r drd\theta. The integral \int_{\theta=0}^{\pi/2}\int_{r=0}^{2\cos\theta} \sqrt{9-r^2} * r
  • #1
bobsmiters
12
0
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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  • #2
bobsmiters said:
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:
 
  • #3
bobsmiters said:
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
 
  • #4
… one integral at a time …

bobsmiters said:
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:
 
  • #5
[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]
 
  • #6
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?
 
  • #7
You're thinking of spherical coordinates.

dA = dxdy = rdrdtheta
dV = dxdydz = r^2sin(theta)dr*dtheta*dphi
 
  • #8
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.
 
  • #9
Vid said:
You're thinking of spherical coordinates.

dA = dxdy = rdrdtheta
dV = dxdydz = r^2sin(theta)dr*dtheta*dphi

Indeed I am! :smile:
bobsmiters said:
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:
 

1. What is a volume integral with two intersecting shapes?

A volume integral with two intersecting shapes is a mathematical calculation used to find the total volume of a three-dimensional region that is formed by the intersection of two different shapes.

2. How is the volume integral calculated for two intersecting shapes?

The volume integral for two intersecting shapes is calculated by setting up a triple integral, where the limits of integration are determined by the boundaries of the intersecting shapes. The integrand is a function that represents the volume of each small element within the region.

3. What is the difference between a volume integral and a surface integral?

A volume integral calculates the total volume of a three-dimensional region, while a surface integral calculates the total surface area of a two-dimensional region. Volume integrals are used to find the volume of a solid object, while surface integrals are used to find the surface area of a curved surface or a solid object.

4. What are some applications of volume integrals with two intersecting shapes?

Volume integrals with two intersecting shapes have many practical applications, such as calculating the volume of a tank, determining the mass of an object with varying density, or finding the volume of a biological structure like a cell or blood vessel.

5. Are there any limitations to using volume integrals with two intersecting shapes?

While volume integrals with two intersecting shapes are a powerful tool for finding the volume of complex regions, the calculations can become very difficult and time-consuming for highly irregular shapes. Additionally, the accuracy of the calculation may be affected by the precision of the measurements used to define the boundaries of the intersecting shapes.

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