Calculating the Area of a Tetrahedron with Double Integral Calculus

[Derill03]

Consider the tetrahedron which is bounded on three sides by the coordinate planes and on the fourth by the plane x+(y/2)+(z/3)=1

Now the question asks to find the area of the tetrahedron which is neither vertical nor horizontal using integral calculus (a double integral)? I think they mean the plane

I am not really sure what to do here, any pointers on where to start? The professor never covered how to do areas with double integrals

 

[Mark44]

Here's something that I think will work. Hopefully, you have drawn a sketch of the tetrahedron, with the x, y, and z intercepts identified. The tetrahedron has, of course, four faces. Two of them are vertical, one is horizontal, and one is determined by the plane whose equation you are given.

If you slice the tetrahedron into some number of equal width slices by cuts that are parallel to the x-z plane, you'll get a bunch of roughly triangular slices. The tops of the slices are trapezoids, not rectangles, but I think if the slices are thin enough that won't matter.

What you want to do is add up (i.e., integrate) the areas of the tops of those slices, and a single integral will do the trick.

The tops of the slices, the trapezoids, have areas that are approximately $\Delta A$, where
$\Delta A \approx$(length of the line segment from the x-y trace to the y-z trace) $\Delta y$

I leave it to you to find a formula for the length of a line segment from a point with fixed y value on on the x-y trace to the corresponding point on the y-z trace. The y-z trace is the intersection of the plane with the y-z plane (hint: every point in the y-z plane has something in common with every other point there). Similar for the x-y trace. Also, you need to figure out the range of y values over which you integrate.

 

[Derill03]

This problem has to be done using a double integral and that is where I am stuck

I know to use 1dxdy and the limits of integration, but I am not sure how to get limits?

 

[Mark44]

Why do you think this problem has to be done with a double integral? You don't show this requirement in your problem statement, I don't believe.

using integral calculus (a double integral)?

I'm assuming that last part is your interpretation of how to do this problem.

 

[Derill03]

I e-mailed my prof to get a little clarification and he suggests using the double integral formula for surface area thalooks like this:

sqrt(1+(partial deriv. x)^2+(partial deriv. y)^2)dxdy

Now that i know what kind of double integral to use I am not sure of what region R is going to be? Would it be the right triangle on xy plane?

sorry for any confusion any help is appreciated

 

[Mark44]

The region R has to be the trianular region in the x-y plane.