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Triple integral over a sphere in rectangular coordinates

  • Thread starter Batmaniac
  • Start date
24
0
1. Homework Statement

Evaluate the following integral:

[tex]
\iiint \,x\,y\,z\,dV
[/tex]

Where the boundaries are given by a sphere in the first octant with radius 2.

The question asks for this to be done using rectangular, spherical, and cylindrical coordinates.

I did this fairly easily in spherical and rectangular coordinates, except for the fact that I got two different answers and I can't figure out where I went wrong! That's not a problem though because I can fix that.


3. The Attempt at a Solution

How would I do this problem in rectangular coordinates? My integral would look like this:

[tex]
\int_{0}^{{\sqrt{4-x^2-y^2}}}\int_{0}^{{\sqrt{4-x^2-z^2}}}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx
[/tex]

Which, without some clever transformations and an extremely messy Jacobian calculation, looks unsolvable.
 

Answers and Replies

206
0
I think you need to rethink your bounds on that one...
 
24
0
How does this look then?

[tex]
\int_{0}^{2}\int_{0}^{2}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx
[/tex]
 
24
0
Hmm, MATLAB tells me that's zero.
 
Dick
Science Advisor
Homework Helper
26,258
618
Your dy limit should depend on x.
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
How does this look then?

[tex]
\int_{0}^{2}\int_{0}^{2}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx
[/tex]
That would be over a square in the xy-plane rising up to the sphere.

Projecting the sphere into the xy-plane gives you the quarter circle x2+ y2= 4, with [itex]0\le x\le 2[/itex], [itex]0\le y\le 2[/itex]. You can let x go from 0 to 2 but then, for each x, y ranges from 0 to [itex]\sqrt{4- x^2}[/itex].
 

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