Say there is an acute isosceles triangle (40,70,70 degrees) with a height of 75 units.
I would like to take the surface integral over the function $f(w,l)=w/l$ where $w$ is the width and $l$ is the length.
I'm not sure if my question makes sense, but here's how I attempted to solve this. Please let me know if I'm doing this correctly.
I know that $\sin(20^\circ )=0.5w/l$, so I have a relationship between $l$ and $w$.
Next, I want to integrate:
$$\int_0^{75} w/l = [(2 l\sin(20))/l] dl =2\sin(20)\cdot 75\quad ??$$
Click the image to see it in more detail:
I will use $L$ instead of $l$ to avoid confusion with $1$. As you say, you want to integrate $\int_0^L \frac {\rho}{t \cdot W(x)}dx$ where $W(x)$ is the width at a given position. At the end you say (using radians) $\sin \left(\frac \pi 9\right)=\frac w{2 L}$ so $W(x)=w\frac{L-x}L=2L\sin \left(\frac \pi 9\right)\frac{L-x}L=2\sin \left(\frac \pi 9\right)(L-x)$ and your integral becomes $$\int_0^L \frac {\rho}{2t\sin \left(\frac \pi 9\right)(L-x)}dx$$