Double Integrals over General Regions

Autor:Ku, Yin Bon (Albert)

Double integrals on Type I regions Suppose

is defined on

, a closed and bounded region in

. How can we define the double integral of

over

? First, we consider

to be the type of regions such that the upper boundary is the graph of

and the lower boundary is the graph of

for

is called a Type I region. Then

can be regarded as a "slice" of the solid under the surface i.e. the graph of

over

for a fixed

. Then we can compute the volume of the solid under the surface by integrating these slices for all

from

. Therefore, the double integral is equal to the following iterated integral:

Example: Suppose

on the domain

bounded by the graphs of

and

and the vertical line

. Compute

. Answer: First of all, the graphs of

and

intersect at

, which means that the range of

is from

. Moreover,

. Therefore, we have

and

. The double integral can be computed as follows:

In the applet below, the left panel shows the region

with the graph of

as upper boundary and the graph of

as lower boundary for

. The right panel shows the solid under the graph of

over

. Also, the "slice" of the solid with fixed

is also shown. You can drag the slider to change the value of

Exercise: Let . Compute , where is the region bounded by the graphs of and . (Hint: First, you need to find the intersection points of the two graphs to determine the range of .)

Double integrals on Type II regions Now we consider

to be another type of regions such that the left boundary is the graph of

and the right boundary is the graph of

for

is called a Type II region. Using the similar idea, we can compute the volume of the solid under the graph of

over

by integrating the slices over all

from

and express the double integral as the following iterated integral:

Example: Let

. Compute

, where

is the region bounded by

and

for

. Answer: Since

, we have

and

. The double integral can be computed as follows:

Exercise: Let . Compute , where is the region bounded by , , and . You can plot the boundaries of using the applet below to visualize the region.

Exercise: Consider , where is the triangular region bounded by , , and . (a) Express the double integral as an iterated integral by regarding as a Type II region. (b) Express the double integral as an iterated integral by regarding as a Type I region. (c) Which of the above is easier to compute? Find the value of the double integral.

Solid between two surfaces Given two functions

and

such that

on a region

. We can compute the volume

between the graphs of

and

over the region

by the following double integral:

Example: Find the volume of the solid bounded by

from below and

from above. Answer: First, we need to find the intersection curve of the two given surfaces. By solving the two given equations, we get

In other words, the intersection curve is a circle as shown in the applet below. Let

. The volume

bounded by the two surfaces is as follows:

Since the solid is rotationally symmetric, we can consider the double integral over the the quarter-disk

in the first quadrant, bounded by

and

for

. Therefore, we have

Use the substitution

, we get

Double Integrals over General Regions

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