Suppose that f (x, y) is defined over a rectangular region R given by R: a £ x £ b, c £ y £ d. Now imagine that a network of lines parallel to the x-and y-axis covers R (See figure 1). These lines divide R into small pieces of area DA = Dx Dy. If we number these in some order - DA₁, DA₂, . . . DA _n, then choose a point (x _k, y _k) in each piece DA _k, we can form the following sum.

If f is continuous throughout R, then as the mesh

figure 1

gets finer, Dx and Dy ® 0 and the sum approaches a limit called the double integral of f over R.

figure 2

When f (x, y) is positive, then the double integral of f over a rectangular region R is the volume of the solid prism bounded below by R and above by the surface z = f (x, y).

EXAMPLE 1:

Find the volume under the plane z = 6 - x - y over the rectangular region R: 0 £ x £ 3, 0 £ y £ 2. (See figure 3 and 4)

SOLUTION:

figure 4

If we apply the method of slicing from calculus I, with slices perpendicular to the x-axis, then the volume is

A (x) is the cross-section area at x. For each value of x, we may calculate

So

When you integrate a double integral, integrate the inner integral first, then the outer integral. Now the question is this, what would happen if we would slice it the other way? Let us try it and see what happens.

What does these two volume calculations with iterated integrals have to do with the double integral

Well, if f (x, y) is continuous over R, then the order of integration does not matter.

If f (x, y) is continuous on the rectangular region R: a £ x £ b, c £ y £ d, then

EXAMPLE 2:

Sketch the region of integration and evaluate the integral.

SOLUTION:

EXAMPLE 3:

Sketch the region of integration and evaluate the integral.

SOLUTION:

figure 7

To define the double integral of a function f (x, y) over a bounded nonrectangular region R, we will imagine R to be covered by a rectangular grid. (See figure 7) When we sum up the areas D A = D x D y, we will include in the partial sum only the areas that lie entirely within the region. Choose an arbitrary point (x _k, y _k) in each  D A_k, then we can write the following sum.

 The only difference between this sum and the one for rectangular regions is that D A _k may not cover the entire region R. As the grid becomes finer, the sum approaches the double integral over the region R.

Double integrals over nonrectangular regions have the same algebraic properties.

If f (x, y) is positive and continuous over R, then the volume of the solid region between R and the surface z = f (x, y) is

If R is a region in the xy-plane that is bounded "above" and "below" by the curves y = g ₂ (x) and y = g₁(x) and on the sides by the lines x = 1 and x = b (See figure 8), then

and

figure 8

 If R is the region bounded by the curves x = h ₂ (y) and x = h ₁ (y) and the lines y = c and y = d, then the volume of this solid is the following double integral.

 FUBINI'S THEOREM (STRONGER FORM)

Let f (x, y) be continuous on a region R.

1.

If R is defined by a £ x £ b, g ₁ (x) £ y £ g ₂ (x), with g ₁ (x) and g ₂ (x) continuous on [a, b], then

2.

 If R is defined by c £ y £ d, h ₁ (y) £ x £ h ₂ (y), with h ₁ (y) and h ₂ (y) continuous on [c, d], then

EXAMPLE 4: 

 Sketch the region of integration and evaluate the integral.

 SOLUTION:

EXAMPLE 5:

Integrate

over the triangular region cut from the first quadrant of the xy-plane by the line x + y = 1.

SOLUTION:

figure 10

First we must sketch the region. (See figure 10) I am going to integrate with respect to y first, then with respect to x.

0 £ y £ 1 - x (cuts parallel to the y-axis)

0 £ x £ 1 (cuts parallel to the x-axis)

EXAMPLE 6:

Sketch the region of integration and write an equivalent double integral with the order of integration reversed.

SOLUTION:

x = y - 2 ® y = x + 2

EXAMPLE 7:

Sketch the region of integration and write an equivalent double integral with the order of integration reversed.

SOLUTION:

EXAMPLE 8:

Sketch the region of integration, determine the order of integration, and evaluate the integral.

SOLUTION:

 EXAMPLE 9:

Find the volume of the solid whose base is the region in the xy-plane that is bounded by the parabola y = 4 - x ² and the line y = 3x, while the top of the solid is bounded by the plane z = x + 4.

 SOLUTION:

 I will let 3x £ y £ 4 - x ². I now need to determine the interval for x. I will determine this interval by setting the equations equal to each other and solve for x.

EXAMPLE 10: 

Evaluate the improper integral.

 SOLUTION: