Triple Integral Spherical Coordinates?

  • #1
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I don't think so since it's not a sphere (disk). I have not learned about cylindrical coordinates and Cartesian is just a pain, so I am assuming I am supposed to use polar or something.

Can someone clear up my confusion?

[tex]\int\int\int_E y\,dV[/tex]

where E lies above the plane z=0, under the plane z=y and inside the cylinder [itex]x^2+y^2=4[/itex]

How would you proceed here if you DON'T want to use Cartesian Coordinates?

Casey
 

Answers and Replies

  • #2
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Duuuudddddeeeeeeee........... I don't get it.... I see a cirlcle (the domain of E) and I automatically think polar... but it is 3-D so tha doesn't male sense right? For 3-D you need spherical coordinates, but that doesn't make sense either.

I am thinking I am not supposed to use Cartesians, cause that integrals a pita. But maybe I am wrong; however we usually don't get integrals that require the use of tables (for now).
 
  • #3
Dick
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To find the volume you integrate the height y, over the area in the x-y plane. Use polar coordinates in the x-y plane. That's really cylindrical coordinates, but you don't have to call it that.
 
  • #4
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Hey Dick! I don't have to much experience with this. Do I set up the integral like this:

[tex]\int\int_D\int_0^y y\,dz\,dA[/tex]

where D is the region bounded by x=0 and x^2+y^2=4

thus the integral becomes

[tex] \int_{\theta=\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{r=0}^2\,y^2\,dr\,d\theta[/tex]

and then just plug in what y is in polar coordinates? Or did I F this one up royally?

Thanks :smile:
 
Last edited:
  • #5
Dick
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Not that royally, but it could use some work. Why y^2? If you are integrating y over the area in the x-y plane it's only a double integral over r and theta. And doesn't theta go from 0 to pi? And isn't the area measure in polar coordinates r*dr*dtheta?
 
  • #6
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I am missing something... where did the third integral go then? The y^2 came from integrating y*dz wtr z--->y*z from z=0 to z=y.

Then I though you integrate that over the x-y area?
 
  • #7
Dick
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If you are setting this up as a triple integral, you integrate 1 over the coordinate limits to get the volume. The integral of 1*dz from z=0 to z=y is y.
 
  • #8
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Maybe I should start over here. If I take the Volume and project it onto the x-y plane, I get D, which takes the form of a semi-circle that follows a path from theta=0 to pi at a radial distance 2 from the origin.

Now I want to integrate the function y wrt r and theta? Is that what you are saying?
 
  • #9
Dick
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Yes, that's exactly what I'm saying. You integrate the height of the region over the area of the base.
 
  • #10
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If you are setting this up as a triple integral, you integrate 1 over the coordinate limits to get the volume. The integral of 1*dz from z=0 to z=y is y.

Why am I integrating 1 ? I am looking for its 4-D volume (or whatever it is called) aren't I.
 
  • #11
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If you are setting this up as a triple integral, you integrate 1 over the coordinate limits to get the volume. The integral of 1*dz from z=0 to z=y is y.

[tex]\int\int\int y\,dz\,dr\,d\theta[/tex]

Once more. Is this my integral? I am just confused, now.

I think it's time for bed! Thanks for the help Dick :smile: I will look at it in the a.m. I am getting too flustered right now. I understand every other triple integral assigned to us except this one. Moreover, I can solve this one in Cartesian Coordinates just fine, so long as I have an integral table handy. So I have no idea why this does not make sense to me in Cylindricals.

In Cartesian I have

[tex]\int_x\int_y\int_z\,y\,dz\,dy\,dx[/tex] .... Not sure what the big deal is now.
 
  • #12
Dick
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Why am I integrating 1 ? I am looking for its 4-D volume (or whatever it is called) aren't I.

I'm going to pretend you didn't say that. A volume integral is the integral of 1 over the 3d coordinate limits. Look it up.
 
  • #13
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I'm going to pretend you didn't say that. A volume integral is the integral of 1 over the 3d coordinate limits. Look it up.

What is the definition of the [tex]\int\int\int_E f(x,y,z)\,dV[/tex]

It is not the volume of E

correct? If I wanted the volume of E I would integrate the function

[tex]\int\int\int_E \,dV[/tex]

Correct?

But my problem does not ask me to find the volume of E. It asks me to evaluate

[tex]\int\int\int_E f(x,y,z)\,dV[/tex]

where f(x,y,z) is simply y. Is that not the case here? I am NOT looking for volume (in the three dimensional case)?
 
  • #14
Dick
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What is the definition of the [tex]\int\int\int_E f(x,y,z)\,dV[/tex]

It is not the volume of E

correct? If I wanted the volume of E I would integrate the function

[tex]\int\int\int_E \,dV[/tex]

Correct?

But my problem does not ask me to find the volume of E. It asks me to evaluate

[tex]\int\int\int_E f(x,y,z)\,dV[/tex]

where f(x,y,z) is simply y. Is that not the case here? I am NOT looking for volume (in the three dimensional case)?

Yes. Completely, utterly, right. I somehow thought you were looking for the volume and you're not. Sorry, sorry. Yes, the y^2 is correct. But shouldn't it be y^2/2?
 
  • #15
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I don't think so: if I have [tex]\int\int_D\int_0^y y\,dz\,dA[/tex]

Just taking care of that innermost integral with respect to z I have

[tex]\int_0^yy\,dz=y\int_0^y\,dz=y*z|_0^y=y^2[/tex]

yes?
 
  • #16
Dick
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Yep, you win again. For a fixed x,y y is a constant. You might be better at this than I am.
 
  • #17
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Yep, you win again. For a fixed x,y y is a constant. You might be better at this than I am.

Yessss! I win! Just kidding.:biggrin: Okay so back to bed. I think I'll be able to sleep now :smile:

So back to my proposal:

Now I have integrated the function with respect to height and got y^2. Now I evaluate a double integral over D of y^2 (but in polar terms=(r*sintheta)^2). Something to the affect of:

[tex]\int\int_D\,y^2\,dA=\int_0^\pi \int_0^2\, r^2*sin^2\theta\,* r\,dr\,d\theta[/tex]
 
  • #18
Dick
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That looks fine to me.
 

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