# If f(x) 0 for a x b, then 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). If f(x,y) 0 for all (x,y) in a region D of the xy plane, then the double integral f(x,y) dx dy = D f(x,y) dA D is defined to be the volume under the graph of f(x,y) on the region D. 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 x (b,c)

(a,d) (b,d) 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 . z f(x,y) = k R y x Since the integral is the volume of a rectangular box with dimensions (a,c) (a,d) b a, d c, and k, then the integral (b,c) must be equal to k(b a)(d c). (b,d) Example Find (0,0,1) z f(x,y) = (1 x) (0,1,1) (0,1,0) (1,0,0) x (1 x) dA . Let R be the rectangle [0,1][0,1]. (1,1,0)

y R Since the integral is the volume of half of a unit cube, then the integral must be equal to 1/2. Cavalieris 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 Px planes, then A(x)x is z approximately the volume of 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 solids volume is b x A(x) dx y a Apply Cavalieris principle to find the volume of the following cone: 5 x y y = 12 13

= circle of th g n e l 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 5 x. 12 25 The area of the slice of the cone at x is A(x) = x2 . 576 b 12 25 The volume of the cone is A(x) dx = x2 dx = 25 576 a 0 To apply Cavalieris 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 A(x) = f(x,y) dy y x

(a,c) (b,c) (a,d) (b,d) b Consequently, f(x,y) dA = b d f(x,y) dy A(x) dx = R a a By reversing the roles of x and y, we may write d b f(x,y) dx c dx . c f(x,y) dA = R c a

dy . 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). Example Find 1 1 (x2 + 4y) dA = R 1 R 1 (x2 + 4y) dy dx = 0 1 x2y + 2y2 dx = y=0 1 1 1 (x2 + 2) dx = x3/3 + 2x 1 (x2 + 4y) dA .

Let R be the rectangle [1,1][0,1]. x = 1 = Alternatively, one could find that 14 3 1 0 1 14 (x2 + 4y) dx dy = 3 1 Example Let R be the rectangle [0 , /2] [0 , /2]. Find /2 (sin x)(cos y) dA = R R /2

(sin x)(cos y) dy dx = 0 0 /2 0 (sin x)(cos y) dA . /2 /2 (sin x)(sin y) dx = (sin x) dx = y=0 0 /2 cos x = 1 x=0 /2 /2 (sin x)(cos y) dx dy = 1 Alternatively, one could find that 0 0