A value for each point in my data-set is defined as $$v=xyz$$where $$a\le x\le b$$$$c\le y\le d$$$$e\le z\le f$$ for ease of visualisation, for now I will assume the following: $$a=c=e=0$$$$b=d=f=1$$$$0\le L\le 1$$
Treating $xyz$ as a cube, I am trying to calculate the volume of the cube where $v<L$. I have plotted a visualisation bellow as an aid in understanding my problem, where $L=0.2$.
For example, in the above cube I am trying to compute the volume of the yellow isovolume.
I thought that I might acheive this by setting $xyz=0.2$, and then triple integrating the function as follows: $$\int_a^b\int_c^d\int_e^f(xyz-0.2)\,dz,dy\,dx$$ Which with my limits of $0$ and $1$ gives the following: $$volume=\frac{x^2y^2z^2}{8} - 0.2xyz$$
However, the result of this function is $-0.075$, which is both smaller than expected, and also negative. I am unsure if this behaviour is due to the asymptotes in the function?
Does anyone have any idea how I can acheive my goal, and also point out the errors in my method? It would be much appreciated.
Thank you in advance!
It's vague to me which quantity are you trying to compute.
We want to have $xyz \le L$, that is upon fixing $x, y$ where they are positive, we want $z$ to satisfy $z \le \frac{L}{xy}$.
If you want to compute the regular volume. \begin{align}&\int_0^1\int_0^1\int_0^{\min(\frac{L}{xy},1)}1\,\,\, dzdydx \\&=\int_0^1\int_0^1 \cdot \min\left(\frac{L}{xy},1\right)\,\,dydx\\&=\int_0^1\int_{\min(1, \frac{L}{x})}^1 \min\left(\frac{L}{xy},1\right)\,\,dydx + \int_0^1\int_0^{\min(1,\frac{L}{x})} \min\left(\frac{L}{xy},1\right)\,\,dydx\\ &=\int_0^1\int_{\min(1, \frac{L}{x})}^1\frac{L}{xy}\,\,dydx + \int_0^1\int_0^{\min(1,\frac{L}{x})}1 \,\,dydx \\&=\int_0^1\frac{L}{x}\left(-\ln\left(\min(1,\frac{L}{x} \right) \right)\,dx +\int_0^1\min(1,\frac{L}{x}) \,dx\\ &=\int_L^1\frac{L}{x}\left(-\ln\left(\min(1,\frac{L}{x} \right) \right)\,dx \\&+\int_0^L\min(1,\frac{L}{x}) \,dx+\int_L^1\min(1,\frac{L}{x}) \,dx\\ &=\int_L^1\frac{L}{x}\left(-\ln\left(\frac{L}{x} \right) \right)\,dx +\int_0^L1 \,dx+\int_L^1\frac{L}{x} \,dx\\ &=L\left[\frac12\ln^2 \left(\frac{L}x \right) \right]_L^1 +L+L(-\ln L)\\ &=L\left( \frac{\ln^2 L}2-\ln L+1 \right) \end{align}
Suppose you want to integrate $xyz$ over the region. \begin{align}&\int_0^1\int_0^1\int_0^{\min(\frac{L}{xy},1)}{xyz}\,\,\, dzdydx \\&=\frac{1}2\int_0^1\int_0^1xy \cdot \min\left(\frac{L^2}{x^2y^2},1\right)\,\,dydx\\&=\frac{1}2\int_0^1\int_{\min(1, \frac{L}{x})}^1xy \cdot \min\left(\frac{L^2}{x^2y^2},1\right)\,\,dydx + \frac{1}2\int_0^1\int_0^{\min(1,\frac{L}{x})}xy \cdot \min\left(\frac{L^2}{x^2y^2},1\right)\,\,dydx\\ &=\frac{1}2\int_0^1\int_{\min(1, \frac{L}{x})}^1\frac{L^2}{xy}\,\,dydx + \frac{1}2\int_0^1\int_0^{\min(1,\frac{L}{x})}xy \,\,dydx \\&=\frac{1}2\int_0^1\frac{L^2}{x}\left(-\ln\left(\min(1,\frac{L}{x} \right) \right)\,dx + \frac{1}4\int_0^1x\min(1,\frac{L^2}{x^2}) \,dx\\ &=-\frac{L^2}2\int_L^1\frac{1}{x}\left(\ln\left(\frac{L}{x} \right) \right)\,dx + \frac{1}4\int_L^1\frac{L^2}{x} \,dx+\frac{1}4\int_0^Lx \,dx\\ &=-\frac{L^2}{2}\left[-\frac12 \ln^2 \left( \frac{L}{x}\right) \right]_L^1 + \frac{L^2}4(-\ln L)+\frac{L^2}8\\ &=\frac{(L\ln L)^2}{4} - \frac{L^2\ln L}4+\frac{L^2}8\end{align}