evaluate the double integral (iint_{d} 6 x y d a), where (d) is the triangular region with vertices…

Explanation:

Step1: Find the equations of the lines

The line passing through ((0,0)) and ((1,2)) has the equation (y = 2x). The line passing through ((1,2)) and ((0,3)) has the slope (m=\frac{3 - 2}{0 - 1}=-1), and using the point - slope form (y-2=-1(x - 1)), so (y=3 - x).

Step2: Set up the double - integral

We set up the double - integral as (\int_{0}^{1}\int_{2x}^{3 - x}6xydydx). First, integrate with respect to (y): (\int_{0}^{1}6x\left[\frac{y^{2}}{2}\right]{y = 2x}^{y = 3 - x}dx=\int{0}^{1}3x\left[(3 - x)^{2}-(2x)^{2}\right]dx). Expand the integrand: ((3 - x)^{2}-4x^{2}=9-6x+x^{2}-4x^{2}=9-6x - 3x^{2}). So the integral becomes (\int_{0}^{1}(27x-18x^{2}-9x^{3})dx).

Step3: Integrate with respect to (x)

(\int_{0}^{1}(27x-18x^{2}-9x^{3})dx=\left[\frac{27x^{2}}{2}-6x^{3}-\frac{9x^{4}}{4}\right]_{0}^{1}). Substitute the upper and lower limits: (\frac{27}{2}-6-\frac{9}{4}=\frac{54 - 24 - 9}{4}=\frac{21}{4}).

Answer:

(\frac{21}{4})