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Mathematica · Feb 5, 2009, 04:19 PM

Convert the integral from rectangular coordinates to spherical coordinates.
it is a triple integral.. from left to right

Integ (-2 to 2), integ (sqrt [4-x^2] to -sqrt[4-x^2]), iteg (x^2+y^2) to 4)) integrant X dzdydx

again... I have the answer but I don't know how to get it...
thank you .

even if you just show me the limits and how you get them I really appreciated .
thank you

galactus · Feb 6, 2009, 01:49 PM

I thought I gave a halfway decent description in your first post of this same problem. As it is, when you evaluate it you get 0. We need to set it up a little differently using symmetry to get the volume another way. If that is what you're after? The xy region is a circle of radius 2. Can you see what the z region is? It appears to be paraboloid bounded above by the plane z=4.

$\int_{-2}^{2} \;\ \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \;\ \int_{x^{2}+y^{2}}^{4}x \;\ dzdydx$

Is your solution 128/15?

If you integrate x w.r.t z, you get zx.

Use your first set of limits of evaluate this portion of the triple integral.

$4x-(x^{3}-xy^{2})$

Now, integrate this w.r.t y

$\int_{-2}^{2} \;\ \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}[4x-x^{3}+xy^{2}]dydx$

Continue?

BTW, I see you have the name Mathematica. Do you happen to have Mathematica? Very nice math software.

I doubt if you respond back letting me know if this helps or not anyway.

Mathematica · Feb 9, 2009, 09:07 PM

Hi, thank you for your answer, I have the mathematica program and it is very nice to have it specially when taking some calculus classes. However, the feedback regarding your answer is that the answer is wrong well I don't even know that you gave me the answer or just a hint because our teacher gave us this question as an extra credit problem and she said that she spent about 4 hours solving the problem that is why I put up this problem in here so that somebody can solve it completely.
But anyway thanks for your time .

galactus · Feb 10, 2009, 08:05 AM

I can solve it well enough by hand but the solution, as is, is 0.

Run it through Mathematica and it'll tell you that. I ran it through my TI and that is what it

Gave me as well. What solution are you looking for?

Let me know what your teacher got. They may have made a mistake if it took them 4 hours to do. It's not that bad. Just a triple integral. Nothing fancy.

Mathematica · Feb 12, 2009, 10:44 AM

I am actually looking for limits of that integral before the integration those are the ones that are hard to get

galactus · Feb 12, 2009, 10:49 AM

You posted the limits. Please post the problem exactly as it was presented so I

May help you better. Do you want to convert to polar or spherical or something like that?