Double Integrals: Find Region Between Surface & Triangle

[Rubik]

I am trying to find the region between a surface z= x+4y and the region D in the x-y plane, where the region is the triangle with verticies (1,1) (2,3) (0,0).. However I am not sure how to come up with the double integral?

 

[HallsofIvy]

Okay, so the base of the region is the triangle with vertices (1,1,0), (2,3,0), and(0, 0, 0) (every point in the xy-plane has z-component 0). I would do this as two separate integrals, taking x going from 0 to 1, then from 1 to 2. For x between 0 and 1, y ranges between the line y= x (from (0, 0, 0) to (1, 1, 0)) to the line y= (3/2)x (from (0, 0, 0) to (2,3, 0)). For x between 1 and 2 the upper line is still the same, y= (3/2)x, but the lower line is now the line from (1, 1, 0) to (2, 3, 0), given by y= 2x- 1. That is, the volume is given by
$$\int_{x=0}^1\int_{y= x}^{(3/2)x} z dydx+ \int_{x= 1}^2\int_{y= 2x-1}^{(3/2)x} z dydx$$
$$= \int_{x=0}^1\int_{y= x}^{(3/2)x} (x+ 4y) dydx+ \int_{x= 1}^2\int_{y= 2x-1}^{(3/2)x} (x+ 4y) dydx$$