How would one use calculus II to come up with the surface área of a cube. This is what i have attempted:if a square is defined as $|y|+|x|=L$ where L is the length of the sides and $ds=\sqrt{1+(dy/dx)^2}$. And we know that the surface área equation is: SA=$$\int 2pix*ds$$ When i do this equation with the formula of a square it doesnt give me the surface área of a cube? From 0 to Z of $$\int 2pi*(L-X)*\sqrt{1+1}$$ because $(dy/dx)^2$ of y-x is simply 1. It doesnt give the surface are of a cube? How come and how would i actually find it.

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    $\begingroup$ I don't believe that $|y| + |x| = L$ parametrizes the surface of a cube. Instead you should probably consider each of the 6 sides separately and parametrize them. $\endgroup$ – Osama Ghani May 11 '17 at 10:22
  • $\begingroup$ can you please elaborate more? what do you mean exactly? $\endgroup$ – Sir Smiles May 11 '17 at 10:31
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    $\begingroup$ In the plane, $|x| + |y| = L$ is a diamond with diagonal $2L$. In 3D this would be an infinite cuboid (i.e. rectangular prism), not a cube. Let's do one face of the cube for example, the one in the x-y plane. We can parametrize this by $ 0 \leq x \leq L, 0 \leq y \leq L$. As a function, $y$ is not dependent on $x$. So $ds = \sqrt{1} = 1$. Now the area is $\int_{0}^{L} \int_{0}^{L} dy dx = L^2$. Repeating this for each face, you will get the total area is $6L^2$. $\endgroup$ – Osama Ghani May 11 '17 at 10:37
  • $\begingroup$ @SirSmiles: You're using the formula the surface area swept out by revolving a graph $y = f(x)$ about an axis. First, the cube is not a surface of rotation. Second, if you tried to calculate the surface area of a cylinder (the nearest rotational analogue of a cube) including the disks on the ends, you'd find the cylinder is not swept out by revolving a graph (there are vertical lines at the ends). The point is, formulas are tools that work in specified circumstances. If you don't ensure the preconditions of a formula are met, you can't expect the formula to give accurate information. $\endgroup$ – Andrew D. Hwang May 11 '17 at 14:17

The cubical volume is not a volume original by a rotation. You need a simple cartesian expression: to get the volume of a cube you simply use a triple integration.

You need actually a triple integration of the sides.

Let any corner lie at the origin, and be the length of a side $\ell$

Let's call the sides lying along the three axis as $x, y, z$ (and of course $x = y = z = \ell$).

The infinitesimals are of course $dx, dy, dz$.

Thus it's a simple triple integration from $0$ to $\ell$ for the volume:

$$V = \int\int\int dx\ dy\ dz$$

each integral is to be understood as $\int_0^{\ell}$, therefore:

$$V = x\cdot y\cdot z \bigg|_{0}^{\ell} = \ell\cdot \ell\cdot \ell - 0\cdot 0\cdot 0 = \ell^3$$


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