For A={(x,y)inRR^2; 0<=y<=3, x/10<=y<=x}
intint_Asqrt(xy-y^2)dxdy = int_{y=0}^3(int_{x=y}^{10y}sqrt(xy-y^2)dx)dy
qquad = int_{y=0}^3(2/(3y)(xy-y^2)^(3/2))_{x=y}^{10y}dy
qquad = int_{y=0}^3[2/(3y)(10y^2-y^2)^(3/2)-2/(3y)(y^2-y^2)^(3/2)]dy
qquad = 2/3int_{y=0}^3 27y^2dy=2/3 times 27/3(3^3-0^3)=162
To get some geometric intuition into this, let us take a look at this graphically. The region A is shown shaded below :
The double integral actually means "breaking up" this region into tiny pieces (like the darkened rectangle dx\ dy in the picture below) and summing up the integrand times the area over all such pieces.
We can decide to either integrate over x first, which essentially means adding up over all rectangular pieces at the same y to get the contribution from the horizontal strip shown below :
The limits for this integration is decided by how much x can vary at this fixed value of x. From the geometry, it is clear that x varies from y to 10y - hence the limits.
We now integrate over y, This means summing up over all the horizontal strips, from y=0 to y=3 :
You could have decided to integrate over y first , so that you would have vertical strips first, which are then added over.Let me leave this as an exercise!