# Thread: Help with definite triple integral.

1. ## Help with definite triple integral.

I have to evaluate the following integral;

$\displaystyle \int_{-2}^{-1}\int_{-1}^{-2}\int_{-2}^{-1} x^2/(x^2+y^2+z^2) dx dy dz$

I sort of solved the first part (dx) with the integration formula:
$\displaystyle \int 1/(x^2+a^2) dx = 1/a *tan^{-1}(x/a)$

I did the following:
First I established $\displaystyle (y^2+z^2)=a^2$
$\displaystyle \int_{-2}^{-1} x^2/(x^2+a^2) dx= \int_{-2}^{-1} {1-a^2/(x^2+a^2)} dx= \int_{-2}^{-1} 1 dx -\int_{-2}^{-1} a^2/(x^2+a^2) dx=(-1-(-2)) -a^2*\int_{-2}^{-1} 1/(x^2+a^2) dx$
$\displaystyle =1 -a^2*(1/a *tan^{-1}(x/a))_{-2}^{-1}=1 -a^2*(1/a *tan^{-1}({-2}/a)-1/a *tan^{-1}({-1}/a))=1 -a*tan^{-1}({-2}/a)+ a*tan^{-1}({-1}/a))$
Then substitute a back to it's original values:
$\displaystyle 1 -\sqrt{(y^2+z^2)}*tan^{-1}({-2}/\sqrt{(y^2+z^2)})+ \sqrt{(y^2+z^2)}*tan^{-1}({-1}/\sqrt{(y^2+z^2)}))$

After this I have no idea what to do.

Any help would be appreciated but if possible I would like to know how to solve the entire problem.

2. ## Re: Help with definite triple integral.

Originally Posted by jollybenito
I have to evaluate the following integral;

$\displaystyle \int_{-2}^{-1}\int_{-1}^{-2}\int_{-2}^{-1} x^2/(x^2+y^2+z^2) dx dy dz$
Where did this integral come from? Is it an attempt to solve some problem? If so, could you state the original problem for us?

3. ## Re: Help with definite triple integral.

It isn't from a real life problem, I have been interested in taking a post graduate course in Science Data, it's from the page of sample questions and it's the only one I haven't been able to solve, I wanted to ask specifically cause I have always had difficulty with integrals that involve trigonometric functions and/or division.
I annex a picture from the original question, it only states "Calculate the following integral".
Don't really know how to deal with it, any hint, help or reference to a similar problem would be highly appreciated.

4. ## Re: Help with definite triple integral.

Let

$\displaystyle I=\int _{-2}^{-1}\int _{-2}^{-1}\int _{-2}^{-1}\frac{x^2}{x^2+y^2+z^2}dxdydz$

$\displaystyle J=\int _{-2}^{-1}\int _{-2}^{-1}\int _{-2}^{-1}\frac{y^2}{x^2+y^2+z^2}dydxdz$

$\displaystyle K=\int _{-2}^{-1}\int _{-2}^{-1}\int _{-2}^{-1}\frac{z^2}{x^2+y^2+z^2}dzdydx$

now, $I=J=K$