If f ( x )  0 for a  x  b , then b f ( x ) dx a

1. If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition of this integral as the area is typically done by defining the integral as a limit of sums; this leads to the Fundamental Theorem of Calculus, which states that such integrals can be calculated using anti-differentiation. The definition of this integral can then be extended so that f(x) may possibly take on negative values (and areas below the x axis will be negative).

2. If f(x,y)  0 for all (x,y) in a region D of the xy plane, then the double integral is defined to be the volume under the graph of f(x,y) on the region D. f(x,y) dx dy = f(x,y) dA DD When the region D is a rectangle R = [a,b][c,d], then the volume is that of the following solid: z z = f(x,y) (a,c) y (a,d) (b,c) x (b,d)

3. Example Suppose f(x,y) = k where k is a positive constant, and let R be the rectangle [a,b][c,d]. Find f(x,y) dA . R z f(x,y) = k y x Since the integral is the volume of a rectangular box with dimensions b– a, d– c, and k, then the integral must be equal to (a,c) (a,d) (b,c) k(b– a)(d– c). (b,d) (1 – x) dA . R Example Let R be the rectangle [0,1][0,1]. Find z f(x,y) = (1 – x) (0,0,1) (0,1,1) Since the integral is the volume of half of a unit cube, then the integral must be equal to (0,1,0) y 1/2. (1,0,0) (1,1,0) x

4. Cavalieri’s Principle : Suppose a solid is sliced by a series of planes parallel to the yz plane, each labeled Px , and the solid lies completely between Pa and Pb . Let A(x) be the area of the slice cut by Px . If x represents the distance between any pair of parallel planes, then A(x)x is approximately the volume of Px z A(x) the portion of the solid between Px and its successor. Summing the approximations A(x)x is an approximation to the volume of the solid. Taking the limit of the sum as x  0, we find that the solid’s volume is b A(x) dx a x y

5. Apply Cavalieri’s principle to find the volume of the following cone: 5 — x 12 y y = length = 13 circle of radius 2.5 x length = 12 z For 0  x  12, the slice of the cone at x parallel to the yz plane is a circle with diameter The area of the slice of the cone at x is A(x) = 5 — x . 12 25 —–x2 . 576 b 12 The volume of the cone is A(x) dx = 25 —–x2dx = 576 25 a 0

6. To apply Cavalieri’s principle to the double integral of f(x,y) over the rectangle R = [a,b][c,d], we first find, for each value of x, the area A(x) of a slice of the solid formed by a plane parallel to the yz plane: z z = f(x,y) d f(x,y) dy c A(x) = y (a,c) (a,d) x (b,c) (b,d) bd f(x,y) dy dx . ac b A(x) dx = a Consequently, f(x,y) dA = R By reversing the roles of x and y, we may write db f(x,y) dx dy . ca f(x,y) dA = R

7. These equations turn out to be valid even when f takes on negative values. The integrals on the right side of each equation are called iterated integrals (and sometimes the brackets are deleted). (x2 + 4y) dA . Example Let R be the rectangle [–1,1][0,1]. Find R 1 1 1 1 (x2 + 4y) dA = (x2 + 4y) dydx = x2y + 2y2dx = R y = 0 –1 0 –1 1 Alternatively, one could find that 1 14 — 3 (x2 + 2) dx = x3/3 + 2x = 1 1 x = –1 – 1 14 — 3 (x2 + 4y) dxdy = 0 –1

8. Example Let R be the rectangle [0 , /2] [0 , /2]. Find (sin x)(cos y) dA . R /2 /2 (sin x)(cos y) dA = (sin x)(cos y) dydx = R 0 0 /2 /2 /2 /2 (sin x)(sin y) dx = (sin x) dx = – cos x = 1 y = 0 x = 0 0 0 /2 /2 (sin x)(cos y) dxdy = 1 Alternatively, one could find that 0 0