I was assigned an exercise in which I have to calculate a double integral of a given function, in an area which is made from four lines. $D$ is defined by the relations: $x+y=1$ , $x-y=1$ , $x+y=3$ , $x-y=-1$ . The integral is $$\iint\limits_D (x+y)^3 \cdot (x-y)^2 \,\mathrm{d}x\,\mathrm{d}y$$
I came to the conclusion that the lines form a rotated square, with edges the points $A(0,1)$ $B(1,2)$ $C(2,1)$ $D(1,0)$.
I guess my last step is to calculate the double integral in that area.
My problem is, can I split the rotated square in 2 isosceles' triangles with points $ABC$, $ADC$ and sum those double integrals together, for more ease? If so, do I have to mention a theorem of some sort that I can't remember?
Is there another way to calculate the double integral on that area?
Thanks in advance for your help!
On the one hand, brute-forcing without a transformation we have $$ \iint_D (x+y)^3(x-y)^2\,dxdy $$ $$ = \int_0^1\int_{1-y}^{y+1}(x+y)^3(x-y)^2\,dxdy + \int_1^2\int_{y-1}^{3-y}(x+y)^3(x-y)^2\,dxdy $$ $$ \stackrel{\text{power rule}}{=} $$ $$\frac{2}{15}\left(\int_0^1 8 y^6 + 20 y^3 + 15 y^2 + 15y\, dy + \int_1^2-8 y^6 + 20 y^3 + 585 y^2 - 1449 y + 910\,dy\right) $$ $$ \stackrel{\text{power rule}}{=}\frac{87}{35}+\frac{439}{105} = \frac{20}{3} $$On the other hand, using the substitution $u=x+y,v=x-y$, whose Jacobian is $2$, we have $$ \iint_D (x+y)^3(x-y)^2\,dxdy $$ $$ =\frac{1}{2} \int _1^3 \int _{-1}^{1} u^3 v^2 \,dvdu = \frac{20}{3} $$