Chapter Twelve Multiple Integrals

Calculus Section 12.1 Double Integrals Over Rectangles Goals Goals Volumes and double integrals Volumes and double integrals Midpoint Rule Midpoint Rule Average value Average value Properties of double integrals Properties of double integrals

Calculus Volumes and Double Integrals Given a function f (x, y), defined on a closed rectangle Given a function f (x, y), defined on a closed rectangle Suppose: f(x, y) ≥ 0. Question: What is the volume of the solid S under the graph of f and above R? Question: What is the volume of the solid S under the graph of f and above R?

Calculus Volumes (cont’d)

Calculus Volumes (cont’d) We do this by We do this by dividing the interval [a, b] into m subintervals [x i-1, x i ] of equal width  x = (b – a)/m and dividing the interval [a, b] into m subintervals [x i-1, x i ] of equal width  x = (b – a)/m and dividing [c, d] into n subintervals [y j-1, y j ] of equal width  y = (d – c)/n. dividing [c, d] into n subintervals [y j-1, y j ] of equal width  y = (d – c)/n. Next we form the subrectangles Next we form the subrectangles each with area  A =  x  y :

Calculus Volumes (cont’d)

Calculus Volumes (cont’d) We choose a sample point We choose a sample point Then we can approximate the part of S that lies above each R ij by a thin rectangular box with base R ij and height Then we can approximate the part of S that lies above each R ij by a thin rectangular box with base R ij and height The volume of this box is the height of the box times the area of the base rectangle: The volume of this box is the height of the box times the area of the base rectangle:

Calculus Volumes (cont’d)

Calculus Volumes (cont’d) Following this procedure for all the rectangles and adding the volumes of the corresponding boxes, we get an approximation to the total volume of S: Following this procedure for all the rectangles and adding the volumes of the corresponding boxes, we get an approximation to the total volume of S: This is illustrated on the next slide: This is illustrated on the next slide:

Calculus Volumes (cont’d)

Calculus Volumes (cont’d) As m and n become larger and larger this approximation becomes better and better. As m and n become larger and larger this approximation becomes better and better. Thus we would expect that Thus we would expect that We use this expression to define the volume of S. We use this expression to define the volume of S.

Calculus Double Integral Limits of the preceding type occur frequently in a variety of settings, so we make the following general definition: Limits of the preceding type occur frequently in a variety of settings, so we make the following general definition:

Calculus Double Integral (cont’d) A volume can be written as a double integral: A volume can be written as a double integral:

Calculus Double Integral (cont’d) The sum in our definition of double integral is called a double Riemann sum and is an approximation to the double integral. The sum in our definition of double integral is called a double Riemann sum and is an approximation to the double integral. If f happens to be a positive function, then the double Riemann sum is the sum of volumes of columns and approximates the volume under the graph of f. If f happens to be a positive function, then the double Riemann sum is the sum of volumes of columns and approximates the volume under the graph of f.

Calculus Example Estimate the volume of the solid that lies Estimate the volume of the solid that lies above the square R = [0, 2]  [0, 2] and above the square R = [0, 2]  [0, 2] and below the elliptic paraboloid z = 16 – x 2 – 2y 2. below the elliptic paraboloid z = 16 – x 2 – 2y 2. Divide R into four equal squares and choose the sample point to be the upper right corner of each square R ij. Divide R into four equal squares and choose the sample point to be the upper right corner of each square R ij. Sketch the solid and the approximating rectangular boxes. Sketch the solid and the approximating rectangular boxes.

CalculusSolution The squares are shown on the next slide. The squares are shown on the next slide. The paraboloid is the graph of f(x, y) = 16 – x 2 – 2y 2 and the area of each square is 1. Approximating the volume by the Riemann sum with m = n = 2, we have The paraboloid is the graph of f(x, y) = 16 – x 2 – 2y 2 and the area of each square is 1. Approximating the volume by the Riemann sum with m = n = 2, we have

Calculus Solution (cont’d)

Calculus Solution (cont’d) Thus 34 is the volume of the approximating rectangular boxes shown: Thus 34 is the volume of the approximating rectangular boxes shown:

Calculus Using More Squares We get better approximations to the volume in the preceding example if we increase the number of squares. We get better approximations to the volume in the preceding example if we increase the number of squares. The next slides show how the columns start to look more like the actual solid when we use 16, 64, and 256 squares: The next slides show how the columns start to look more like the actual solid when we use 16, 64, and 256 squares:

Calculus Using More Squares (cont’d)

Calculus Using More Squares (cont’d)

Calculus The Midpoint Rule We use a double Riemann sum to approximate the double integral. We use a double Riemann sum to approximate the double integral. The sample point to be the center The sample point to be the center

Calculus Example Use the Midpoint Rule with m = n = 2 to estimate the value of Use the Midpoint Rule with m = n = 2 to estimate the value of Solution We evaluate f(x, y) = x – 3y 2 at the centers of the four subrectangles shown on the next slide: Solution We evaluate f(x, y) = x – 3y 2 at the centers of the four subrectangles shown on the next slide:

Calculus Solution (cont’d)

Calculus Solution (cont’d) The area of each subrectangle is ΔA = ½, so The area of each subrectangle is ΔA = ½, so

Calculus Using More Subrectangles If we keep dividing each subrectangle into four smaller ones, we get the Midpoint Rule approximations shown. These values approach the exact value of the double integral, –12.

Calculus Average Value We define the average value of a function f of one variable defined on a rectangle R as We define the average value of a function f of one variable defined on a rectangle R as where A(R) is the area of R. If f(x, y) ≥ 0, the equation If f(x, y) ≥ 0, the equation

Calculus Average Value (cont’d) says that the box with base R and height f ave has the same volume as the solid that lies under the graph of f. If z = f(x, y) describes a mountainous region and we chop off the tops of the mountains at height f ave, then we can use them to fill in the valleys so that the region becomes completely flat: If z = f(x, y) describes a mountainous region and we chop off the tops of the mountains at height f ave, then we can use them to fill in the valleys so that the region becomes completely flat:

Calculus Average Value (cont’d)

Calculus Properties of Double Integrals On the next slide we list three properties of double integrals. On the next slide we list three properties of double integrals. We assume that all of the integrals exist. We assume that all of the integrals exist. The first two properties are referred to as the linearity of the integral: The first two properties are referred to as the linearity of the integral:

Calculus Properties (cont’d) If f(x, y) ≥ g(x, y) for all (x, y) in R, then If f(x, y) ≥ g(x, y) for all (x, y) in R, then

Calculus Review Volumes and double integrals Volumes and double integrals Definition of double integral using Riemann sums Definition of double integral using Riemann sums Midpoint Rule Midpoint Rule Average value Average value Properties of double integrals Properties of double integrals