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Unformatted text preview: (9/28/10) Math 20C. Lecture Examples (Revised) Sections 15.1 and 15.2. Double integrals A region R in the xy-plane is bounded if it can be enclosed in a sufficiently large circle. Its boundary is piecewise smooth if it consists of a finite number of graphs y = y ( x ) or x = x ( y ) of functions that are defined on finite closed intervals and have continuous first derivatives on those intervals. A partition , R 1 ,R 2 , .. ., R N (1) of such a region is created by slicing it with piecewise smooth curves. Figure 2 shows a partition of the region R of Figure 1 into four subregions. To measure the size of a subregion we use its diameter , which is the diameter of the smallest circle that contains it. FIGURE 1 FIGURE 2 The interior of a region is obtained by removing any boundary points from it. The closure is obtained by adding any boundary points that are not in it. A function z = f ( x,y ) is piecewise continuous on a region if there is a partition of the region such that in the interior of each subregion, f is equal to a function that is continuous on the closure of the subregion. Definition 1 (Riemann sums) Suppose that z = f ( x,y ) is piecewise continuous on a bounded region R with a piecewise-smooth boundary and that (1) is a partition of R . A Riemann sum for integraldisplay integraldisplay R f ( x,y ) dxdy corresponding to the partition (1) is a sum of the form N summationdisplay j =1 f ( x j ,y j ) [Area of R j ] (2) where for each j = 1 , 2 , . .. ,N , ( x j , y j ) is a point in R j where f is defined. Double integrals over such regions are the limits of Riemann sums: Definition 2 (Double integrals) If z = f ( x,y ) is piecewise continuous on a bounded region R with a piecewise-smooth boundary, then the double integral integraldisplay integraldisplay R f ( x,y ) dxdy is the limit of Riemann sums (2) as the number of subintervals in the partition tends to infinity and their diameters tend to zero. Lecture notes to accompany Sections 15.1 and 15.2 of Calculus, Early Transcendentals by Rogawski 1 Math 20C. Lecture Example Solutions (9/28/10). Sections 15.1 and 15.2, p. 2 For the region R of Figure 1, the partition of Figure 2, and the positive function z = f ( x, y ) of Figure 3, the Riemann sum is equal to the total volume of four prism-shaped solids as in Figure 4, whose sides are vertical, whose bases are the four subregions of the partition, and whose horizontal tops intersect the graph of f at x = x j ,y = y j . Because such collections of solids approximate the solid under the graph, we are led to the following definition. [Volume] = integraldisplay integraldisplay R f ( x,y ) dxdy [Volume] = A Riemann sum] FIGURE 3 FIGURE 4 Definition 3 (Volume) If R is a bounded region in an xy-plane with a piecewise-smooth boundary and z = f ( x,y ) is piecewise continuous and has nonnegative values on R , then the volume of the solid above R and below the graph z = f ( x,y ) is integraldisplay integraldisplay R f ( x,y ) dxdy (Figure 3)....
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