How do I use integration to find the surface area of 3x+2y+z=6 in the first octant, without relying on a 3D graphing calculator?

Answer 1

We can use Heron's formula.

The three vertices are (2, 0, 0), (0, 3, 0), and (0, 0, 6)

Thus, the three sides have lengths sqrt(13), sqrt(40), and sqrt(45)

Then, the area is sqrt(s(s-a)(s-b)(s-c)) = 11.2497,which turns out to equal 3 sqrt(14)

Another way to solve:

Using the segment for (2, 0, 0) to (0, 3, 0) as the base, the vector is (2, -3,0)

Then, points on this segment are of the form (x,3 - 3x/2, 0), and the vector from this to (0, 0, 6) is

(-x,3x/2-3,0)

The 2 vectors must be perpendicular. Thus (-x,3x/2-3,0) . (2, -3, 0) = -2x -9x/2 + 9 = 0

Thus, 9 = 13x/2

x = 18/13 y = 3 - 3x/2 = 3 - 27/13 = 12/13

Then, the area is 1/2 b h

b = sqrt(2^2+3^2) = sqrt(13)

h = sqrt(6^2+(12/13)^2+(18/13)^2) = sqrt(504/13)

A = 1/2 sqrt(13)sqrt(504/13) = sqrt(126) = 3 sqrt(14)