Mathematica Mathematica Returning Conditional Expression

AI Thread Summary
The discussion centers on performing a volume integral in Mathematica, where the user encounters an issue with the output being a ConditionalExpression instead of a straightforward result. The integral is structured in three parts over specified limits, and while the returned answer is mathematically correct, it includes conditions related to complex numbers. The user seeks guidance on how to instruct Mathematica to focus solely on real numbers to avoid these conditions in the output. It is noted that in Mathematica 8, the Assumptions option can be used with the Integrate function to specify that only real numbers should be considered, which may resolve the issue.
davidallen353
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I am attempting to do a simple volume integral over a particular shape. However Mathematica will not simply return an answer. For my input

Code:
Volume = Integrate[
   Integrate[Integrate[1, {z, 0, c}], {y, 0, b}], {x, 0, a}] + 
  Integrate[
   Integrate[Integrate[1, {z, c, c + a}], {y, b/2, b - \[Rho]}], {x, 
    0, a}] + 
  Integrate[
   Integrate[
    Integrate[
     1, {z, c, a + c - Sqrt[-b^2 + 2 b*y - y^2 + \[Rho]^2]}], {y, 
     b - \[Rho], b}], {x, 0, a}]

It returns

Code:
ConditionalExpression[
 a b c + a^2 (b/2 - \[Rho]) + 
  a (a \[Rho] - 
     1/4 \[Pi] \[Rho] Sqrt[\[Rho]^2]), ((Im[\[Rho]] >= 0 && 
      Im[b] Re[\[Rho]] <= Im[\[Rho]] Re[b]) || (Im[b] Re[\[Rho]] >= 
       Im[\[Rho]] Re[b] && Im[\[Rho]] <= 0)) && 
  Re[\[Rho] Im[b] - b Im[\[Rho]]]^2/Re[\[Rho]]^2 <= 1]

The answer it actually returns is correct, so my question is how do I tell Mathematica that I am only concerned with Real numbers. I am using Mathematica 8
 
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Assumptions is an option you can supply to Integrate
 

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