0
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.