/ 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.
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 θ, φ, r
x = r cos θ sin φ
y = r sin θ sin φ
z = r cos φ
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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/ 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)