I'm currently working on the evaluation of the volume and lateral surface area of solid such as the one shown below:
In which the nonlinear faces in the cross-sections can be described by any unknown functions $x=f(z)$ and $y=g(z)$ (which may not be the same), with symmetry with respect to the $z$ axis. The illustrations below can help to clarify what I mean.
The only boundary conditions I have a priori for the functions are (assuming only the right side given the symmetry):
$$f(z=0)=L/2$$
$$g(z=0)=B/2$$
My question is on how can I write the expressions for the volume and lateral surface area as functions of $f(z)$ and $g(z)$? If the solid was made by the extrusion of a unique cross-section it would be easier since I could evaluate the curve length/area and multiply by the depth, but since it has 2 different cross-sections I don't know how to deal with the area/volume of the corners.
I figured it out by simple geometry for the case in which the functions are linear, i.e., the solid is similar to a pyramid frustum but with different inclinations on every two sides. Now I need to find a more general expression for any function so that, later on, I could find these functions by minimizing a functional problem. Note that the problem is posed as a function of the $z$ coordinate and not the $x$ or $y$ as usual since I have the boundaries determined only for this axis: $z\in[0,H]$.
P.S.: Given the symmetry, the problem can be posed as a quarter of the solid shown above as well.
Thank you for your attention. Any help provided would be truly appreciated!
Best Answer
Let's consider a slab of thickness $dz$ at some height $z$. The width and length of the slab are $2f(z)$ and $2g(z)$. Then the volume of the slab is $$dV=(2f(z))(2g(z))dz=4f(z)g(z)dz$$ Then the volume of the solid is obtained by integrating $dV$: $$V=4\int_0^Hf(z)g(z)dz$$ Now let's look at the area of one of the sides, say the one that you see when you look along $y$ axis. We divide this area into rectangles, with the long side along $x$ axis, so the width is $2f(z)$. If the surface would be vertical when seen along $x$ axis, then the height of the rectangle would be just $dz$. But in general it is not. So we would need to replace $dz$ by some $dl$, a length element along $g(z)$. This is given by $$dl=dz\sqrt{ 1+\left(\frac{dg(z)}{dz}\right)^2}$$ So the area of this side is going to be $$A_1=\int_0^H2f(z)\sqrt{ 1+\left(\frac{dg(z)}{dz}\right)^2} dz$$ You can compute then for every face, to get $$A=4\int_0^H\left(f\sqrt{1+g'^2}+g\sqrt{1+f'^2}\right)dz$$