Double Integral of f(x,y) in a Triangle Domain

[quasar987]

I am asked to calculate the double integral of the function

f(x,y) = (2x+3y)^2 = 4x^2 + 12xy + 9y^2

on the domain defined by a triangle whose summits(?) are at (-1,0), (0,1) and (1,0). I chose to integrate from left to right. So the bounds of my integral are

\int_0^1 \int_{y-1}^{1-y} (4x^2 + 12xy + 9y^2)dxdy

 

[arildno]

1) Vertices, not summits
2) Looks good to me

 

[wais]

To calculate the double integral of f(x,y) in a triangle domain, we first need to set up the bounds of the integral. In this case, the triangle domain is defined by the points (-1,0), (0,1), and (1,0), so we can set up the bounds of the integral as follows:

- The outer integral will be with respect to y, and it will go from 0 to 1 since those are the y-coordinates of the points at the top and bottom of the triangle.
- The inner integral will be with respect to x, and it will go from the left side of the triangle to the right side. Since we are integrating from left to right, the left side of the triangle will be given by the equation x = y-1, and the right side will be given by the equation x = 1-y.

Plugging these bounds into the double integral, we get:

\int_0^1 \int_{y-1}^{1-y} (4x^2 + 12xy + 9y^2)dxdy

Now, we can use the properties of double integrals to split this into two separate integrals:

\int_0^1 \int_{y-1}^{1-y} 4x^2dxdy + \int_0^1 \int_{y-1}^{1-y} 12xydxdy + \int_0^1 \int_{y-1}^{1-y} 9y^2dxdy

From here, we can easily integrate each of these individual integrals using the power rule for integration. This will give us:

\int_0^1 \left[ \frac{4x^3}{3} + 6x^2y + 9xy^2 \right]_{y-1}^{1-y} dy

Simplifying this further, we get:

\int_0^1 \left[ \frac{4(1-y)^3}{3} + 6(1-y)^2y + 9(1-y)y^2 - \frac{4(y-1)^3}{3} - 6(y-1)^2y - 9(y-1)y^2 \right] dy

Now, we can simplify the expressions inside the brackets and combine like terms:

\int_0^1 \