New here, so I'll try get this asked right the first go.
I have a double integral I need solved. I actually have the solution, and have verified it myself through numerical integration, but I would like to see the process myself that was used to get the solution. Essentially I have a circle that is split into 3 sections. Lets call them top bottom and middle. The Top and bottom are equal. What am trying to do is calculate the size of each of these portions. I'm not too concerned about the formulas within the integral, they should be right (for those curious, I'm attempting to find these areas in order to multiply them by the stress associated in each piece, sum the moments about the centre, top+bottom times stress +middle times stress = 0, but I digress)
$h=$ distance to centre of circle to the centroid of the top or bottom half.
$r=$ radius of circle
$x =$ horizontal distance
$z =$ vertical distance.
$s=$ stress (a constant)
$P=$ Peak Strain (a constant)
$R=$ Residual Strain (a constant) (not little $r$)
$E =$ Modulus of Elasticity (a constant)
The piece of the integral I'm stuck on (bottom and top portions) is below. When I try solving it I end up with $\arctan (\text{something} /0)$ .
$$\int_{-h}^{-r} \int_{\sqrt{r^2-z^2}}^{-\sqrt{r^2-z^2}} \left[s-E\frac{z}{r}(p-R)\right]z \,dx\,dz$$
Any help solving this would be fantastic. As I said I have a spreadsheet that solves it quite accurately numerically but I would love to know what steps I need to solve it using integration. (I've been playing with polar coordinates but have gotten myself lost)
Thanks!! and I hope this obeys all the forum rules.
Why don't you try switching the variables of integration (integrate wrt $z$ before $x$)? For the bottom section, at $z = -h$, this intersects the circle at $x = -\sqrt{r^2-h^2}$ and $x = \sqrt{r^2-h^2}$, both constants. Let's call them $-x_0$ and $x_0$ for simplicity
In the $z$ direction this corresponds the bottom half to the circle $z = -\sqrt{r^2-x^2}$ to the line $z = -h$. In the $x$ direction $-x_0 \le x \le x_0$ so we have this integral
$$ \int_{-x_0}^{x_0} \int_{-\sqrt{r^2-x^2}}^{-h} \left[sz - \frac{E(p-R)}{r}z^2\right] \, dz\,dx$$
That's one less square root to work with. Try it and let me know what you get