Double Integration: Non-Rectangular Domains

Transcription

1 Double Integration: Non-Rectangular Domains Thomas Banchoff and Associates June 18, Introduction In calculus of one variable, all domains are intervals which are subsets of the line. In calculus of two variables, there is a richer collection of domains; in addition to rectangular domains, we have, for example, sectors of circles defined in polar coordinates, and regions bounded by two function graphs in the plane. 2 Domains in Polar Coordinates One example of a non-rectangular domain is a sector obtained when we describe the plane in polar coordinates and restrict ourselves to a r b c θ d. The area of a sector with radius between r i and r i+1 and angle between θ j and θ j+1 is approximated by r i (r i+1 r i )(θ j+1 θ j ). So, the total area of our sector is the integral of the function f(r, θ) = r with respect to r and θ, where r goes from a to b and θ goes from c to d. If we want to calculate the volume above a sector in polar coordinates and under the graph of a function g in cylindrical coordinates, then we approximate 1

2 by the volume of a collection of prisms defined over subsectors, each with constant height given by the value of the function at the lowest values of r and θ in the subsector. The approximating sum is then identical to the approximating sum for the value of the double integral of the function rg(r, θ) defined over the rectangle a r b, c θ d. The volume of a function graph in cylindrical Cordinates. START : caption The graph of r 2 theta with approximating prisms. END : caption This demonstration is similar to Demonstration 1 in Lab 5, but it shows the approximating prisms for a function in cylindrical coordinates and the volume of those approximating prisms. If this sector has variable density g(r, θ), for instance, then the total mass can be approximated by integrating the function rg(r, θ). Exercise 1 Examine the effect of changing the number of subdivisions in the calculation of the volume under the graph of z = cos(2θ). Exercise 2 Estimate the total mass of a semicircular region of radius 2 in the upper half plane (so 0 r 2, 0 θ π, where the density at each point is r cos 2 (θ) ). Exercise 3 Find the volume under the graph of z = e r2 over the disc of radius 5. 2

3 3 Planar Domains Bounded By Function Graphs Another sort of domain in the xy-plane is the region between the graphs of two functions d(x) defined over an interval a x b. An approximation of this domain can be obtained by subdividing the interval a x b into n subintervals, and then subdividing each of the intervals c(x i ) y d(x i ) into m subintervals. This will determine a collection of (m 1)(n 1) rectangles forming an approximation of the region. In order to compute the volume lying under the graph of a function f(x, y) defined over such a domain, we first approximate by small rectangles in this manner, and then calculate the total volume of the prisms with these rectangles as bases, each with height equal to the value of the function at the lower lefthand corner of the subrectangle. The volume under a graph over a region bounded by function graphs d(x). This demonstration is very similar to the second demo in lab 5. In the Function Graph window, we display the function graph between the two curves with the approximating prisms. We have added the Volume as a Function of n and m window in order to study the effect of increased divisions on the approximation of the integral. Note that these n and m are independent of the subdivisions you have specified for displaying the function graph. The volume displayed in the input window is for the displayed function graph with its subdivisions. As n and m become large, the volume of the prisms more and more closely approximates the volume under the surface. Note that you only need to specify one number for the y subdivisions since the max and min are set by the values of d(x). Exercise 4 Estimate the volume under the graph of z = 1 + x 2 + y 2 over the region between y = x 2 and y = x 3, 0 x 1. 3

4 Exercise 5 Estimate the volume under the graph of z = cos(e x+y ) above the region between y = cos(x) and y = x 2 ( π 2 )2. We may also compute such a volume by means of repeated integration. The fundamental principle is that the volume under the graph of f over the region between d(x) from a to a particular value x 0 of x is a function F (x) of the value of x, and the derivative F (x 0 ) of this function with respect to x is the area under the slice curve f(x 0, y) as y goes from c(x 0 ) to d(x 0 ). Volume as a function of x. START : caption Left: Domain defined by the curves c(x)=x+1 and d(x)=x with the subdomain shown in blue. Right: The function graph of x 2 y 2 with prisms to x 0 over the same subdomain.end : caption In this demonstration, for a chosen value x 0 of x, we indicate the portion of the domain in the xy-plane and between the graphs of d(x) from a to x 0, and we also indicate the corresponding region on the graph of the function. We show the approximating prisms over that region of the domain. Note that when you turn the prisms on you should also turn the domain on, because the prisms are hard to interpret without it. We also display the graph of the area under the slice curve f(x 0, y) where y goes from c(x 0 ) to d(x 0 ) as a function of x (that is, as a function of the choice of x 0 ) as well as the graph of the integral of this function (this being the function F (x) ). We print the value of F (x 0 ), i.e. the integral of the area function from 4

5 a to x 0. Recall that when the slice curve is below the z = 0 plane, the area between the plane and the graph is counted as negative area. Exercise 6 Analyze the integrals in the last two exercises from the point of view of a the area under the slice curve at x = t as t goes from a to b. Similarly to a domain that could be described as a region bounded by graphs of functions of x, if a domain in the plane can be described as the region between the graphs of a(y) and b(y) as y ranges from c to d, then we may approximate the region by rectangles by first subdividing the interval c y d into m subintervals, then subdividing each interval a(y j ) x b(y j ) into n subintervals. We may then calculate the area of the region as before, and we may find the volume under the graph of f and above the region. The volume under a graph over a region bounded by function graphs a(y) and b(y). START : caption Left: The function graph x 2 y 2 with approximating prisms. Right: A view of the volume as a function of n and m. From this view you will NOT be able to understand how close the approximation is.end : caption This demonstration is analogous to Demonstration 2, showing a domain defined as a region between two functions of y as y ranges from c to d. 5

6 Exercise 7 Evaluate the volume of the function z = e x2 over the domain 0 x 1, x y 1. Exercise 8 Evaluate the volume under the graph of z = 1 + x 2 + y 2 over the region between x = y 1/3 and x = y 1/2. For this type of domain we could define a volume function F (y) analogous to the one defined above with respect to x. For domains in the plane that can be defined as either regions between two functions of x or regions between two functions of y, we may compute the volume under a function graph over the domain either by integrating the area under the slice curve f(x 0, y) where y goes from c(x 0 ) to d(x 0 ) with respect to x from x 0 = a to x 0 = b, or by integrating the area under the slice curve f(x, y 0 ) where x goes from a(y 0 ) to b(y 0 ) with respect to y from y 0 = c to y 0 = d. 6