Find the volume between the planes z = 2x+3y +6 and z=2x + 7y +8, and over the triangle vertices (0,0), (3,0), and (2,1)

Answer 1

The line from (0, 0) to (2, 1) is y = 2x

The line from (3,0) to (2, 1) is y = 3 - x

Using I[a,b] as the integral from a to b and E[a,b] as the evaluation from a to b

To make this a single set of integrals, you need to define x in terms of y on the triangle.

Define the line segment from (0, 0) to (2, 1) as x = 1/2 y and the line segment from (3, 0) to (2, 1) by x = 3 - y

Note how y ranging from 0 to 2 will then define the triangle.

I[0, 2]I[1/2 y, 3 - y]I[2x+3y-8,2x+7y+8] dz dx dy =

I[0, 2]I[1/2 y, 3 - y] z E[[2x+3y-8,2x+7y+8] dx dy =

I[0, 2]I[1/2 y, 3 - y] 4y+2 dx dy =

I[0, 2] (4y+2) x E[1/2 y, 3 - y] dy =

I[0, 2] (4y+2)(3 - 3/2y) dx =

3 I[0, 2] (2y + 1)(2 - y) dy =

3 I[0, 2] - 2y^2 + 3y + 2 dy =

3 (-2/3y^3 + 3/2y^2 + 2y E[0, 2] =

3(-16/3 + 6 + 4) = 14

Is there an easier way to do the problem? Realize that we are finding the volume of 4y+2 over the triangle. That will just be 2(area of triangle) + 4 (area of triangle)(y centroid)

The y centroid for this triangle is 2/3

Then, 2(3) + 4(3)(2/3) = 6 + 8 = 14