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Computing Double Integrals and Volumes with Maple


*****

Exercises on Double Integrals

Iterated integrals are straight forward in Maple. Here are some examples,

> Int(Int(x^4-y^2,x=1..2),y=0..1) = int(int(x^4-y^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)=
> 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.

Problem1:


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);

picture a picture here


Solution1:


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


Problem2:


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.

Solution2:


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);

picture a picture here

The actual surface on the first octant looks like this,

> plot3d(z2,x=-1..2,y=-1..2,axes=frame);

picture a picture here

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=x2) 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

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>
Last modified: Tue Nov 5 09:47:29 EST 1996