Having a bit of a problem calculating the volume of a take-away box:

I originally wanted to use integration to measure it by rotating around the x-axiz, but realised that when folded the top becomes a square, and the whole thing becomes rather irregular. Since it differs in circumference I won’t be able to measure it like I planned.

Is there any method or formula that can be used to measure a shape like this, or do I just have to approximate a cylinder and approximate a box and add those two together?

**Answer**

A different approximation could be to take the cross-section at each height not as a linear interpolation between top and bottom surface, but as squares with rounded corners. This fits the photograph more closely, as the linear interpolation approach has a discontinuity in the radius of curvature at the bottom: The point that at the top is an edge of the square is modelled a crease along the entire side, while in the photograph, no such crease can be seen.

If we take the cross-section to have the above shape of a rounded square, we can calculate its area by using the formulas for circles and rectangles:

$$A = \pi \cdot r^2 + 2 \cdot a \cdot b – b^2$$

Using $b=a-2\cdot r$, we can eliminate that variable:

$$A = a^2 + (\pi -4) \cdot r^2$$

Now we assume that those variables vary linearly between top and bottom: $r$ goes from $r_0$ at the bottom to $0$ at the top, and $a$ goes from $2r_0$ at the bottom to $a_0$ at the top, giving us:

$$r(h) = \left(1-\frac{h}{h_0}\right)\cdot r_0$$

$$a(h) = \left(1-\frac{h}{h_0}\right)\cdot 2r_0 + \frac{h}{h_0} \cdot a_0$$

and thus:

$$A(h) = \left(1-\frac{h}{h_0}\right)^2 \cdot \pi \cdot r_0^2 + \left(1-\frac{h}{h_0}\right) \cdot \frac{h}{h_0} \cdot 4 \cdot r_0 \cdot a_0 + \left(\frac{h}{h_0}\right)^2 \cdot a_0^2$$

or, better ordered for integration:

$$A(h) = \pi \cdot r_0^2 + \frac{h}{h_0} \cdot \left( 4\cdot r_o \cdot a_0 – 2 \cdot \pi \cdot r_0^2 \right) + \left(\frac{h}{h_0}\right)^2 \cdot \left( a_0^2 – 4\cdot r_0 \cdot a_0 + \pi \cdot r_0^2\right) $$

Which gives us then:

$$ V = \int_0^{h_0} A(h) \mathrm{d}h= h_0 \cdot \pi \cdot r_0^2 + \frac{1}{2} \cdot \frac{h_0^2}{h_0} \cdot \left( 4\cdot r_o \cdot a_0 – 2 \cdot \pi \cdot r_0^2 \right) + \frac{1}{3}\cdot\frac{h_0^3}{h_0^2} \cdot \left( a_0^2 – 4\cdot r_0 \cdot a_0 + \pi \cdot r_0^2\right)$$

$$=h_0\cdot\left(\frac{1}{3}\pi r_0^2 + \frac{2}{3} r_0 a_0 + \frac{1}{3} a_0^2\right)$$

And here is a quick render of what this looks like in 3d:

This result can be obtained classically without any integrals as well, using only the formulas for the volume of conic solid and for cuboids. To do this, we split the solid into nine parts:

The central, yellow part is simply a square pyramid, and has a volume of $\frac{1}{3}\cdot h_0\cdot a_0^2$. The four magenta pieces are oblique cones with a quarter-circle as base. Again, using the formula for conic solids, their volume is each $\frac{1}{12}\cdot h_0\cdot\pi\cdot r_0^2$

And the four cyan pieces are irregularly shaped tetrahedrons, but we can determine their volume by adding a few pieces to see how they fit into a cuboid of the measurements $a_0\times r_0 \times h_0$:

In the more general case of $\frac{a_0}{2} \neq r_0$, there will be a fourth green piece needed (which would go in front and block our view of everything). However, the green pieces together form a prisma of height $a_0$ and a top surface area of $\frac{1}{2} \cdot r_0 \cdot h_0$, and the two blue-grey pieces are two pyramids with height $h_0$ and rectangular bases of size $\frac{1}{2} a_0 \times r_0$. Therefore, the volume of the tetrahedron is:

$$a_0 \cdot h_0 \cdot r_0 – a_0 \cdot \frac{1}{2} \cdot r_0 \cdot h_0 – 2 \cdot \frac{1}{3} \cdot h_0 \cdot \frac{1}{2} a_0 \cdot r_0 = \frac{1}{6} h_0 \cdot r_0 \cdot a_0$$

Putting it all together, we get the volume as:

$$V=\frac{1}{3}\cdot h_0\cdot a_0^2 + 4\cdot \frac{1}{12}\cdot h_0\cdot\pi\cdot r_0^2 + 4 \cdot \frac{1}{6} h_0 \cdot r_0 \cdot a_0$$

$$=h_0\cdot\left(\frac{1}{3}\pi r_0^2 + \frac{2}{3} r_0 a_0 + \frac{1}{3} a_0^2\right)$$

A more smooth model along the same lines would be to interpolate superellipses between the bottom and top, which similarly would give a creaseless change in the radius of curvature beneath the corners. However, superellipse areas have a formula involving the gamma function, and are therefore not easy to integrate again to get a volume.

**Attribution***Source : Link , Question Author : Nemui , Answer Author : AlienAtSystem*