Transcript 11.5 Double Integration

Double
Integration
Greg Kelly, Hanford High School, Richland, Washington
f  x, y   4  x  y
Find the volume under this surface between 0<x<2
and 0<y<1.

f  x, y   4  x  y
z
0  x  2 and 0  y  1
We can sketch the graph by
putting in the corners where
(x=0, y=0), (x=2, y=0),
(x=0, y=1), (x=2, y=1).
x
y

z
We could hold x constant and
take a slice through the shape.
The area of the slice is given by:
1
 4  x  y dy
0
x
y
The volume of the slice is
area . thickness
 area  dx

We can add up the volumes of
the slices by:
z
2
1
0
0
  4  x  y dy dx
1
1 2
0 4 y  xy  2 y 0dx
2
1
0 4  x  2dx
2
x
y
2
1 2 1
4x  x  x
2
2 0
8  2 1
5

The base does not have to be a rectangle:
f  x, y   3  x  y
with triangular base between the x-axis, x=1 and y=x.
y
1

x
0 0
3  x  y dy dx
area of slice
x
slice
thickness
of slice
Add all slices from 0 to 1.