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triple integral in spherical coordinates

Posted on 2006-10-28
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Consider the following triple integral:

     / 2pi   / pi       / sqrt(2)
    |        |           |               f(p,x,y) p^2 sin y dp dx dy
    |        |           |
   / 0      / 3pi/4   /0

Where p is rho, x is phi, y is theta in spherical coordinates. I'm trying to convert this integral to cartesian coordinates.

Its rather easy to see that the outer most integral will have limits of (-sqrt(2) to sqrt(2)) with respect to x, the middle integral will have limits of (-sqrt(2 - x^2) to sqrt(2 - x^2)) with respect to y, my problem is the inner most integral with respect to z. The book has the answer as something like -sqrt(2 - x^2 - y^2) to sqrt(x^2 + y^2). I'm just not seeing where this is coming from.


Thanks.
Brian
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Question by:BrianGEFF719
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by:ozo
ID: 17827989
/ pi
 |
 |
 / 3pi/4
is a cone shaped spherical sector
so I would have expected the limits to be from sqrt(x^2 + y^2) to sqrt(2 - x^2 - y^2)
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jkmyoung earned 500 total points
ID: 17869185
since phi > pi/2 z is negative.
Since the radius is at most sqrt 2, this means that x² + y² + z² <= 2
z² <= 2 - x² - y²
-(2 - x² - y²) <= z <= 2 - x² - y²

Now for the restriction based on phi.

Note the conversion from &#952;, &#966;, r
x = r cos &#952; sin &#966;
y = r sin &#952; sin &#966;
z = r cos &#966;

cos (3pi/4) = -sqrt(2)        
sin (3pi/4) = sqrt(2)
cos (pi) = -1
sin (pi) = 0

so z must be at most -sqrt(x² + y²)
z <= -sqrt(x² + y²)
and since
x² + y² <= 2
<=> (2 - x² - y²) >= 0
=> -sqrt(x² + y²) <= (2 - x² - y²)

-(2 - x² - y²) <= z <= - (x² + y²)
I'm assuming you left the negative sign out of your orignal question.
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