Riemann Sum Comparison

Transcription

1 Double Integrals Riemann Sum Comparison Riemann sum to approximate area Subdivide [a, b] into n intervals I.

2 Double Integrals Riemann Sum Comparison Riemann sum to approximate area Subdivide [a, b] into n intervals I. Interval width: x i = x i x i 1. Choose sample point xi in each I The n Riemann f (xi ) x i sum i=1 approximates area under curve.

3 Double Integrals Riemann Sum Comparison Riemann sum to approximate area Subdivide [a, b] into n intervals I. Interval width: x i = x i x i 1. Choose sample point xi in each I The n Riemann f (xi ) x i sum i=1 approximates area under curve. Integral is limit of Riemann sums. n f (x) dx = lim f (xi ) x i a max x i 0 i=1 (If exists) b

4 Double Integrals Riemann Sum Comparison Riemann sum to approximate area Riemann Sum to approximate volume (Today: Rectangles [a, b] [c, d].) Subdivide [a, b] into n intervals I. Interval width: x i = x i x i 1. Choose sample point xi in each I The n Riemann f (xi ) x i sum i=1 approximates area under curve. Integral is limit of Riemann sums. n f (x) dx = lim f (xi ) x i a max x i 0 i=1 (If exists) b Subdivide [a, b] into m intervals. Subdivide [c, d] into n intervals.

5 Double Integrals Riemann Sum Comparison Riemann sum to approximate area Riemann Sum to approximate volume (Today: Rectangles [a, b] [c, d].) Subdivide [a, b] into n intervals I. Interval width: x i = x i x i 1. Choose sample point xi in each I The n Riemann f (xi ) x i sum i=1 approximates area under curve. Integral is limit of Riemann sums. n f (x) dx = lim f (xi ) x i a max x i 0 i=1 (If exists) b Subdivide [a, b] into m intervals. Subdivide [c, d] into n intervals. Each subrect. [x i 1, x i ] [y j 1, y j ] has area A ij = x i y j. The m n Riemann f (xi, yi ) A ij i=1 j=1 sum approximates volume under surface.

6 Double Integrals Riemann Sum Comparison Riemann sum to approximate area Riemann Sum to approximate volume (Today: Rectangles [a, b] [c, d].) Subdivide [a, b] into n intervals I. Interval width: x i = x i x i 1. Choose sample point xi in each I The n Riemann f (xi ) x i sum i=1 approximates area under curve. Integral is limit of Riemann sums. n f (x) dx = lim f (xi ) x i a max x i 0 i=1 (If exists) b Subdivide [a, b] into m intervals. Subdivide [c, d] into n intervals. Each subrect. [x i 1, x i ] [y j 1, y j ] has area A ij = x i y j. The m n Riemann f (xi, yi ) A ij i=1 j=1 sum approximates volume under surface. The double integral is limit of R.S. f (x, y) da = lim R.S. R max x i 0 y i 0 (If exists)

7 Double Integrals Notes and an example If f is continuous, the limit always exists!

8 Double Integrals Notes and an example If f is continuous, the limit always exists! The volume under the surface is approximated by adding up the volumes of a bunch of rectangular prisms.

9 Double Integrals Notes and an example If f is continuous, the limit always exists! The volume under the surface is approximated by adding up the volumes of a bunch of rectangular prisms. Example. Estimate the volume that lies above the square R = [0, 2] [0, 2] and below the surface z = 16 x 2 2x 2 by calculating a Riemann sum with four terms.

10 Double Integrals Notes and an example If f is continuous, the limit always exists! The volume under the surface is approximated by adding up the volumes of a bunch of rectangular prisms. Example. Estimate the volume that lies above the square R = [0, 2] [0, 2] and below the surface z = 16 x 2 2x 2 by calculating a Riemann sum with four terms. Answer: The region R can be divided into four squares, each with area A =.

11 Double Integrals Notes and an example If f is continuous, the limit always exists! The volume under the surface is approximated by adding up the volumes of a bunch of rectangular prisms. Example. Estimate the volume that lies above the square R = [0, 2] [0, 2] and below the surface z = 16 x 2 2x 2 by calculating a Riemann sum with four terms. Answer: The region R can be divided into four squares, each with area A =. Choose a sample point in each region. (Ex: Upper right corner.)

12 Double Integrals Notes and an example If f is continuous, the limit always exists! The volume under the surface is approximated by adding up the volumes of a bunch of rectangular prisms. Example. Estimate the volume that lies above the square R = [0, 2] [0, 2] and below the surface z = 16 x 2 2x 2 by calculating a Riemann sum with four terms. Answer: The region R can be divided into four squares, each with area A =. Choose a sample point in each region. (Ex: Upper right corner.) Volume f (1, 1) A 11 + f (1, 2) A 12 + f (2, 1) A 21 + f (2, 2) A 22 =

13 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums.

14 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums. Instead, we use the method of iterated integrals: Calculate the volume by adding slices together.

15 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums. Instead, we use the method of iterated integrals: Calculate the volume by adding slices together. For a fixed x, we can find the area of an (infinitesimal) slice of f (x, y) by integrating over y.

16 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums. Instead, we use the method of iterated integrals: Calculate the volume by adding slices together. For a fixed x, we can find the area of an (infinitesimal) slice of f (x, y) by integrating over y. Then we can integrate these areas over all slices to give the volume.

17 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums. Instead, we use the method of iterated integrals: Calculate the volume by adding slices together. For a fixed x, we can find the area of an (infinitesimal) slice of f (x, y) by integrating over y. Then we can integrate these areas over all slices to give the volume. Area of slice: A(x) = y=d y=c f (x, y) dy [Think of x as constant.]

18 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums. Instead, we use the method of iterated integrals: Calculate the volume by adding slices together. For a fixed x, we can find the area of an (infinitesimal) slice of f (x, y) by integrating over y. Then we can integrate these areas over all slices to give the volume. y=d Area of slice: A(x) = f (x, y) dy [Think of x as constant.] x=b y=c x=b [ y=d ] Volume: V = A(x) dx = f (x, y) dy dx. x=a x=a y=c

19 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums. Instead, we use the method of iterated integrals: Calculate the volume by adding slices together. For a fixed x, we can find the area of an (infinitesimal) slice of f (x, y) by integrating over y. Then we can integrate these areas over all slices to give the volume. y=d Area of slice: A(x) = f (x, y) dy [Think of x as constant.] x=b y=c x=b [ y=d ] Volume: V = A(x) dx = f (x, y) dy dx. We write x=a b d a c f (x, y) dy dx. x=a y=c

20 Double Integrals Calculating double integrals We never calculate a double integral by taking limits of Riemann sums. Instead, we use the method of iterated integrals: Calculate the volume by adding slices together. For a fixed x, we can find the area of an (infinitesimal) slice of f (x, y) by integrating over y. Then we can integrate these areas over all slices to give the volume. y=d Area of slice: A(x) = f (x, y) dy [Think of x as constant.] x=b y=c x=b [ y=d ] Volume: V = A(x) dx = f (x, y) dy dx. We write x=a b d a c f (x, y) dy dx. x=a y=c The order of the dy and dx tells you which to integrate first. Work from the inside out.

21 Double Integrals Fubini s Theorem If f is continuous on the rectangle R = [a, b] [c, d] then R f (x, y) da = b d a c f (x, y) dy dx = d b c a f (x, y) dx dy

22 Double Integrals Fubini s Theorem If f is continuous on the rectangle R = [a, b] [c, d] then R f (x, y) da = b d a c f (x, y) dy dx = d b c a f (x, y) dx dy Also true in general if f is bounded on R f discontinuous only on a finite number of smooth curves the iterated integrals exist

23 Double Integrals Fubini s Theorem If f is continuous on the rectangle R = [a, b] [c, d] then R f (x, y) da = Also true in general if f is bounded on R b d a c f (x, y) dy dx = d b c a f (x, y) dx dy f discontinuous only on a finite number of smooth curves the iterated integrals exist Intuition: In our picture, we could have sliced the volume by fixing y instead of x.

24 Double Integrals Fubini s Theorem If f is continuous on the rectangle R = [a, b] [c, d] then R f (x, y) da = Also true in general if f is bounded on R b d a c f (x, y) dy dx = d b c a f (x, y) dx dy f discontinuous only on a finite number of smooth curves the iterated integrals exist Intuition: In our picture, we could have sliced the volume by fixing y instead of x. Take away message: When f is nice, we can choose the order of integration to make our life easier.

25 Double Integrals Double integrals Example. Find R y sin(xy) da where R = [1, 2] [0, π].

26 Double Integrals Double integrals Example. Find R y sin(xy) da where R = [1, 2] [0, π]. Is it easier to integrate with respect to.

27 Double Integrals Double integrals Example. Find R y sin(xy) da where R = [1, 2] [0, π]. Is it easier to integrate with respect to. Properties of double integrals

28 Double Integrals Double integrals Example. Find R y sin(xy) da where R = [1, 2] [0, π]. Is it easier to integrate with respect to. Properties of double integrals When f (x, y) is a product of (a fcn of x) and (a fcn of y) over a rectangle [a, b] [c, d], then the double integral decomposes nicely: [ b ] [ d ] g(x)h(y) da = g(x) dx h(y) dy R a c

29 Double Integrals Double integrals Example. Find R y sin(xy) da where R = [1, 2] [0, π]. Is it easier to integrate with respect to. Properties of double integrals When f (x, y) is a product of (a fcn of x) and (a fcn of y) over a rectangle [a, b] [c, d], then the double integral decomposes nicely: [ b ] [ d ] g(x)h(y) da = g(x) dx h(y) dy R For general regions R: R (f + g) da = R f da + R g da a c (not necessarily rectangles)

30 Double Integrals Double integrals Example. Find R y sin(xy) da where R = [1, 2] [0, π]. Is it easier to integrate with respect to. Properties of double integrals When f (x, y) is a product of (a fcn of x) and (a fcn of y) over a rectangle [a, b] [c, d], then the double integral decomposes nicely: [ b ] [ d ] g(x)h(y) da = g(x) dx h(y) dy R For general regions R: R (f + g) da = R f da + R g da R cf da = c R f da a c (not necessarily rectangles)

31 Double Integrals Double integrals Example. Find R y sin(xy) da where R = [1, 2] [0, π]. Is it easier to integrate with respect to. Properties of double integrals When f (x, y) is a product of (a fcn of x) and (a fcn of y) over a rectangle [a, b] [c, d], then the double integral decomposes nicely: [ b ] [ d ] g(x)h(y) da = g(x) dx h(y) dy R For general regions R: R (f + g) da = R f da + R g da R cf da = c R f da a c (not necessarily rectangles) If f (x, y) g(x, y) for all (x, y) R, then R f da g da.