# Thread: Surface Area of a Regular Tetrahedron

1. ## Surface Area of a Regular Tetrahedron

How to find the surface area of a regular tetrahedron using calculus? The surface area should be √3 a^2.

I tried to do the same approach that MarkFL did for the volume but I could not figure out how to do it with the area!

2. ## Re: Surface Area of a Regular Tetrahedron

$$\int\int xydS = \int\int xydS_1 + \int\int xydS_2+\int\int xydS_3 + \int\int xydS_4$$

Since you are told the tetrahedron is regular, you know

$$\int\int xydS_1 = \int\int xydS_2 = \int\int xydS_3 = \int\int xydS_4$$

This means, you only need to calculate the area of one side and multiply by 4. Assume $S_1$ is bounded by these lines:

$$x=0$$
$$y=\sqrt{3}x$$
$$y=\sqrt{3}(a-x)$$

So, the area would be:

$$\int\int xy dS_1 = \int_0^{a/2}\sqrt{3}xdx + \int_{a/2}^a \sqrt{3}(a-x)dx = \dfrac{\sqrt{3}a^2}{4}$$

Multiplying by 4 to get the complete area gives $\sqrt{3}a^2$ as desired.

3. ## Re: Surface Area of a Regular Tetrahedron

Thanks a lot SlipEternal. Integrating each triangle alone is a nice way, but I need to integrate the whole regular tetrahedron at once going from 0 to h (height).

We did that method for the volume of a solid regular tetrahedron and it worked fine since we had mass everywhere inside the object.

For the area, I think it is very complicated to do that since there is no mass inside the object! But I am wondering, is it possible to set an integral from 0 to h?

Why would I want to do that?

I am trying to find the mass moment of inertia of a hollow regular tetrahedron.
I can do that only if I have a slice of the height. Then integrating this slice from 0 to the complete height!

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