Surface area of irregualar shape How to find the surface area of an irregular shape like this

 A: If this is built up of cubes with side length $1$, and all cubes are placed exactly along the integer coordinate grid, then it's just a matter of counting the number of cube faces that are visible, and since each face has area $1$, that's it. Be systematic and make sure you haven't left any sides out. That is the easiest way.
One simplification to the manual counting is possible, though: For each face facing in one direction, there is a corresponding face behind it facing the opposite direction. So you don't have to count the faces facing left, for instance, because there are equally many facing right, so you can just count all the faces facing right and double the result.
A: If you really want to try to be clever, you can make this shape such that the number of blocks stacked in each axis is monotonic by removing, adding, displacement, etc. Then it is obvious that the new surface area equals twice of sum of areas of three-view diagrams. Then you try to find the difference of the new shape and the original one.
This is still counting actually, but in a way that occasionally quicker. Just count if counting is possible.
