Question

Find the area of the surface: The part of the plane 2x-5y+z=10 that lies above the triangle with vertices (0,0), (0,6), and (4,0).

Answer 1

I already have the integral done, it's just a matter of finding the bounds.

\[Solve for z: z=10-2x-5y\]
\[dx/dz = -2, dy/dz = -5\]
\[L=\sqrt{(-2)^{2}+(-5)^{2}+1}=\sqrt{30}\]
\[\int\limits_{\int\limits_{D}^{?}}^{?} \sqrt{30} dA\]

Answer 2

I'm just having trouble finding the bounds for y for the double integral. I'm already assuming that the bounds for X will be [0,4], so Y would be [0,some value]. I can take it from there, just need help getting the top Y bound

Answer 3

doesnt y move between 2 lines? define the lines

Answer 4

well its 2 lines, one of them is the y axis ... draw out the region

Answer 5

Like this?

Answer 6

thats a terrible drawing of a triangle region ....