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Double Integrals

Double Integrals

Integrals determine the area under a function. If the function “spreads” between two variables, then a surface is described. The space under this surface has a volume. To determine the volume under the surface you use a double integral.

You can detect the x-axis (right) and the y-axis (left). Between the boundaries a-b and c-d lies the region R. The volume between this bottom and the “ceiling”  is the volume V which can be evaluated by the double integral. Imagine you have a quader with a “curly hat”.

A double integral must have two variables, mostly x and y. The surface is the graph of the two-variable function  to be integrated. The rectangular region R at the bottom of the body is the domain of integration.

Double integral as volume under a surface z = x² − y² (Source: Oleg Alexandrov)

The region under a surface may be of any shape. The simplest case is a rectangle. This will be spoken in this chapter. The next chapter deal with other shapes of the region.

Formula of the double integral

To evaluate a double integral is relatively easy. You begin from the inside variable and treat the other variable as a constant. Normally there is dx and dy, so you begin with x. The R is the abbreviation of region. It means evaluate the integral between a wanted x-boundary and a wanted y-boundary. This letter is nothing but a general indicator for a double integral. The real values must be put in. Outside the y-boundary and inside the x-boundary as you can see at the differentials .

  Example 1

Evaluate the volume under the curve . The region R is  and .

, a beautiful hammock

Put in the boundaries for x and y:

First integrate x. Let y be a constant.

Second, integrate this term with y and put in the y-boundaries:

The volume V is:

Please accustom to the way of integrate the double integral. I prefer the way from inside to outside. It would be possible and right to change this direction, but soon confusion would be inevitable. Fom inside to outside with the given differential variables will avoid mistakes.

  Example 2

Evaluate the double integral of 1. This is a cube.

The double integral of 1 is simply the area of R.

If the region is  and , then

Compare the well-known formula for a cube .

If width = 1, then: