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Using Mathematica I want to solve an integral over a function that contains both determined and undetermined parts which looks like this:

r[\[Theta], z] = Sqrt[(h/2)^2 - z^2] + g[\[Theta]] - h/2

As you can see the h/2 and root terms are determined, but g remains an undetermined function of $\theta$. Now I take an integral over this function to both z and theta:

V = Integrate[[Integrate[1/2 r[\[Theta], z]^2, {z, -h/2, h/2}], {\[Theta], 0, 2 \[Pi]}]

which results in:

$$V= \int_0^{2 \pi } \left(\frac{1}{8} (\pi -4) h^2 g(\theta )+\frac{1}{2} h g(\theta )^2+\frac{1}{48} (10-3 \pi ) h^3\right) \, d\theta$$

As you can see, the last term contains no dependence on $\theta$ and therefore I would like that part of the integral solved. In particular I would like my result to look like this $$V=\frac{1}{24} (10-3 \pi )\pi h^3 + \frac{1}{8} (\pi -4) h^2 \int_0^{2 \pi }g(\theta )\, d\theta+\frac{1}{2} h \int_0^{2 \pi }g(\theta )^2 \, d\theta$$

Is there some way to force mathematica to solve the parts of the function that are not dependent (or with known dependence) on the variable towards which I integrate?

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Yes, there is such a way. Try this, for example. First define a function to integrate;

int[expr_] := Integrate[expr, {z, -h/2, h/2}, {\[Theta], 0, 2 \[Pi]}]

Then map this function onto the terms of your expression:

r[\[Theta], z] = Sqrt[(h/2)^2 - z^2] + g[\[Theta]] - h/2;

Map[int, Expand[1/2 r[\[Theta], z]^2]]

This results in:

$$\int_0^{2 \pi } -\frac{1}{2} h^2 g(\theta ) \, d\theta +\int_0^{2 \pi } \frac{1}{8} \pi h^2 g(\theta ) \, d\theta +\int_0^{2 \pi } \frac{1}{2} h g(\theta )^2 \, d\theta +\frac{1}{24} \pi (10-3 \pi ) h^3 $$

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    $\begingroup$ Great, thanks! Quick comment: did you know that you can copy the output of mathemetica directly as latex by right-clicking on the equation and selecting "Copy as -> Latex"? $\endgroup$ – Michiel Jul 4 '14 at 12:12
  • $\begingroup$ @Michiel No, I did not. $\endgroup$ – Alexei Boulbitch Jul 4 '14 at 14:40

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