Integrate f (x, y) = x^2 + y^2 over the triangular region with vertices (0,0), (1,0), and (0,1). use a definite double integral and show all the steps.

the integral result is I = 1/6

Step-by-step explanation:

For the region with vertices (0,0), (1,0), and (0,1) we have the

boundaries y=1-x , x=0 and y=0 for the integral then

1)  integrating over the region y=1-x and y=0 for y , and then from x=1 to x=0

I = ∫∫ f (x,y) dx*dy = ∫₀¹∫₀¹⁻ˣ  (x^2 + y^2 ) dy*dx =  ∫₀¹ [(1-x)*x^2 + (1/3)(1-x)^3  - 0*x^2 + (1/3)0^3 ] dx = ∫₀¹ [x^2 - (2/3)x^3] dx = [(1/3)x^3 - (1/6) x^4 ]|₀¹=  [(1/3)1^3 - (1/6) 1^4 ] - [(1/3)0^3 - (1/6) 0^4 ] = (1/3) - (1/6) = 1/6

2) integrating over the region x=1-y and x=0 for x , and then from y=1 to y=0 (the same process but changing y for x)