If we let f (x, y) = 1 in the definition of the double integral over a region R, then

Therefore,

EXAMPLE 1:

Sketch the region bounded by the parabola x = y - y ² and the line y = -x. Then find the region's area as an iterated double integral.

SOLUTION:

The upper curve is x = y - y ², and the lower curve is x = -y. Now I must find the points of intersection.

First, I am going to integrate with respect to x, then I will integrate with respect to y. Here is the double integral for the area.

EXAMPLE 2:

Sketch the region bounded by the curves y = ln x and y = 2lnx and the line x = 3 in the first quadrant. Then find the region's area as an iterated double integral.

SOLUTION:

I am going to integrate with respect to y first, then x second. y will go between ln x and 2ln x, and x will go between 1 and e.

EXAMPLE 3:

Given

sketch the region, and then find the area of the region.

SOLUTION:

The average value of an integrable function of a single variable on a closed interval is the integral of the function over the interval divided by the length of the interval. For an integrable function of two variables defined on a closed and bounded region that has measurable area, the average value is

EXAMPLE 4:

Find the average value of f (x, y) = sin (x + y) over the rectangle 0 £ x £ p 0 £ y £ p .

SOLUTION:

The area of the rectangle is p ².

The equations for the center of mass are similar to the equations talked about in calculus I, but with the concept of double integrals, we can accommodate a greater variety of shapes. The mathematical difference between the first moments M _x and M _y and the moments of inertia (second moments), I _x and I _y is that the second moments use the squares of the "lever-arm" distances x and y.

Here are the mass and moment formulas.

DENSITY:

d (x, y)

MASS:

FIRST MOMENTS:

 CENTER OF MASS:

 MOMEMTS OF INERTIA (SECOND MOMENTS)

 ABOUT THE X-AXIS:

ABOUT THE Y-AXIS:

(polar moment)

ABOUT A LINE L:

where r (x, y) equals the distance from (x, y) to L.

RADII OF GYRATION

ABOUT THE X-AXIS:

ABOUT THE Y-AXIS:

ABOUT THE ORIGIN:

I _o is called the polar moment of inertia about the origin and is defined to be I _o = I _x + I _y. The radius of gyration R _x is defined by the equation I _x = MR _x. It tells us how far from the x-axis the entire mass of the plate might be concentrated to give the same I _x. The radius of gyration gives a convenient way to express the moment of inertia in terms of a mass and a length. R _y and R _o are defined in a similar way.

When the density of an object is constant, d , it cancels out of the numerator and denominator when we are calculating the center of mass. So we can let d = 1. When d is constant, the location of the center of mass becomes a feature of the object's shape and not the material of which it is made. We call the center of mass of an object with constant density the centroid of the shape.

EXAMPLE 5:

Find the center of mass of a thin plate of constant density d = 3 bounded by the lines x = 0, y = x, and the parabola y = 2 - x ² in the first quadrant.

SOLUTION:

EXAMPLE 6:

 Find the moment of inertia and radius of gyration about the x-axis of a thin plate bounded by the parabola x = y - y ² and the line x + y = 0 if d (x, y) = x + y.

First, I will find the points of intersection of the two curves.

Now I will find the mass of this thin plate.