Exercises on Double Integrals 
> Int(Int(x^4y^2,x=1..2),y=0..1) = int(int(x^4y^2,x=1..2),y=0..1);
1 2 / /   4 2 88   x  y dx dy =    15 / / 0 1> Int(Int(sin(x+y),x=0..Pi/2),y=0..Pi/2)=
1/2 Pi 1/2 Pi 1/2 Pi / / /      sin(x + y) dx dy =  sin(y) + cos(y) dy    / / / 0 0 0
and this last integral is equal to: 
> Int(int(sin(x+y),x=0..Pi/2),y=0..Pi/2) =
> int(int(sin(x+y),x=0..Pi/2),y=0..Pi/2);
1/2 Pi /   sin(y) + cos(y) dy = 2  / 0
Double integrals can be used to compute volumes. Here are some examples.

Find the volume of the solid lying under the elliptic paraboloid 
> z := 1  (x^2/4 + y^2/9);
2 2 z := 1  1/4 x  1/9 y
and above the square R=[1,1]x[2,2]. The solid looks like this: 
> plot3d(z,x=1..1,y=2..2,axes=frame);

Computing the integral over the square we get: 
> Int(Int(z,x=1..1),y=2..2) = int(int(z,x=1..1),y=2..2);
2 1 / /   2 2 166   1  1/4 x  1/9 y dx dy =    27 / / 2 1
which is approximately, 
> ans1 := evalf(int(int(z,x=1..1),y=2..2),2);
ans1 := 6.1

Find the volume of the solid in the first octant bounded by the surface, 
> z2 := x*sqrt(x^2+y);
2 1/2 z2 := x (x + y)
and the planes x=1, and y=1.

The double integral will compute the volume as the limit of approximations with blocks like the ones shown in the picture below: 
> with(mvcal):
> blockapp(z2,x=0..1,y=0..1);
The actual surface on the first octant looks like this, 
> plot3d(z2,x=1..2,y=1..2,axes=frame);
Computing the double integral we get, 
> Int(Int(z2,x=0..1),y=0..1);
1 1 / /   2 1/2   x (x + y) dx dy   / / 0 0
actually maple had a hard time with this integral in my computer. It refused to compute the exact expression for the integral eventhough a simple substitution (u=x^{2}) does it. Any way after a lot of computing... 
> evalf(");
.4875805666
If we do the integration over x by hand and let maple handle the integration over y we get the exact answer as, 
> ans2 := int((1+y)^(3/2)y^(3/2),y=0..1)/3;
1/2 ans2 := 8/15 2  4/15> evalf(");
.4875805663