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This was just a thought I had while driving this morning, no particular application.
If you make a straight line cut through a solid object such that the resulting two pieces have the same volume, will they have the same surface area? Vice versa?
What about applications to certain shapes? Regular prisms? Logically it seems that it would be true for a sphere considering there's only one way to cut it in half to achieve the same volume.
Note - I'm only considering uniform densities
Join two solids of equal volume by a small (narrow and short) tube - e.g. a sphere and a cube. Cut the tube in half.
The volumes will be equal. The surface area on each side will be very nearly the surface area of the original component solids. Since the surface area does not determine the volume (largest volume for given surface area is s sphere), there is no reason for the two surface areas to be the same.
Once you see this you will realise that equality will only occur in special cases - eg cutting along an axis of symmetry; or by mimicking the proof of the ham sandwich theorem (taking a plane along which to cut the volume in half, and rotating it continuously, and showing by using the intermediate value theorem that there is some plane which will give equal surface areas too).
Not at all. Consider a sphere with a long narrow cylinder of the same volume stuck to it. Cut them at the joint. The cylinder will have much more surface area. Yes, a sphere has enough symmetry to enforce this, and I believe the Platonic solids except the tetrahedron do as well. The tetrahedron fails because you can cut parallel to one face. The volume of a regular tetrahedron is $\frac {a^3}{6\sqrt 2}$ and the area $\sqrt 3 \cdot a^2$. Setting $a$ to $1$ for the original, we cut off half the volume with $a=2^{-\frac 13}$ The new one has a surface of $\sqrt 3 \cdot2^{-\frac 23}\approx 1.091$ The leftover piece has surface $\sqrt 3 -\frac 12 \sqrt 3 \cdot 2^{-\frac 23}\approx 1.186$ because we lost three small triangles and gained one.
Suppose there is a cone of height 3h,slant height 3l & radius r.
Dividing the cone into two equal volumes.ie,cutting at height h from the base & parallel to the base.
So resulting area of top object(which is a cone) is pi*r1*2l(LSA) + Area of base
where
r1^2 = (2l)^2 - (2h)^2 = 4(l^2 - h^2)
so,
r / r1 = 3/2 -> r1 = 2r/3
so, Total surface area of top cone = pi*2r*2l/3(LSA) + Area of base.
Now,area of bottom object = area of top base + area of bottom base + LSA
LSA = LSA of original cone - LSA of top cone = pi*r*3l - pi*2r*3l/3 = pi*r*l
Area of bottom base = pi*r*r
So, comparing both objects' areas,
pi*4*rl/3 + area of base = pi*r*l + area of top base + pi*r*r
area of base of top object = area of top base of bottom object,so both will be cancelled.
4rl/3= rl + r*r
rl/3 = r*r
ie l = 3r
so, the areas will be equal only when l = 3r. For all other cases, areas will be unequal.
Conclusion: For any other object too, there will be a special case where the areas of the resulting objects with equal volume, will be equal, but for all other cases,it'll be unequal.
