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Section 4.1 Double Integrals

4.1 Double Integrals

In calculus I, we learn one of most important reasons for learning calculus, finding areas. We use Riemann sum to define an integral:

let f(x) be a function defined over the interval, [a,b] and we divide the interval into n-subintervals, [xi1,xi] where i=1,...,n, x0=a , and xn=b. We choose xi inside [xi1,xi], then

abf(x)dx=limni=1nf(xi)x

 where x=ban the length of the interval [xi1,xi] (under the assumption that the limit exists). If f(x)0 then abf(x)dx present the area between f(x) and x-axis from x=a to x=b. The idea is built from adding up rectangles with width x=ban and height f(xi) at the interval [xi1,xi]. 

For two variables functions, we have similar definitions and approach. If f(x,y) is a function with two variables defined over R=[a,b]×[c,d], suppose f(x,y)0 for any (x,y) in R. Then we can find the volume under the surface f(x,y) above the xy-plane using Riemann sum. First we divide [a,b] into m-subintervals and [c,d] into n-subintervals and we have the ij-th rectangle [xi1,xi]×[yj1,yj]. We choose (xi,yj) inside [xi1,xi]×[yj1,yj],

then lim(m,n)(,)j=1ni=1mf(xi,yj)xy presents the volume under the surface f(x,y) above the xy-plane ( if the limit exists) where x=bam and y=dcn. 

 

 

Definition: Double integral

Let f(x,y) be a function defined over R. The double integral of the function f(x,y) over the rectangular region R in the xy-plane is defined as

Rf(x,y)dA=lim(m,n)(,)j=1ni=1mf(xi,yj)A.

 

 

Remark:

If f(x,y)0, then the volume V of the solid S, which lies above R in the xy-plane and under the graph of f, is the double integral of the function f(x,y) over the rectangle R. If the function is ever negative, then the double integral can be considered a \textquotedblleft signed\textquotedblright{} volume in a manner similar to the way we defined net signed area.

 

Definition 

Assume a,b,c, and d are real numbers. We define an iterated integral for a function f(x,y) over the rectangular region R=[a,b]×[c,d] as 

cdabf(x,y)dxdy=cd[abf(x,y)dx]dy=cd[A(y)]dy

abcdf(x,y)dydx=ab[cdf(x,y)dy]dx=ab[B(x)]dx.

 

 

Theorem: Fubini’s Theorem 

Suppose that f(x,y) is a function of two variables that is continuous over a rectangular region R=[a,b]×[c,d]. Then the double integral of f over the region equals an iterated integral, 

Rf(x,y)dA=cdabf(x,y)dxdy=abcdf(x,y)dydx

More generally, Fubini\textquoteright s theorem is true if f is bounded on R and f is discontinuous only on a finite number of continuous curves. In other words, f has to be integrable over R.

 

Example 1: Compute the double integral Rf(x,y)dA where f(x,y)=x and R=[0,2]×[1,1].

 

 

 

 

Exercise 1: Compute the double integral Rf(x,y)dA where f(x,y)=y and R=[2,0]×[0,3].

 

 

 

Example 2: Compute the double integral 2002(x+xy)2dxdy.

 

 

 

 

Exercise 2: Compute the double integral0310(yxy)2dxdy.

 

 

 

Remark:

Rf(x)g(y)dxdy=abf(x)dxcdg(y)dy.

 

Example 3: Compute the double integral 1302x2ylnydxdy.

 

 

 

 

Exercise 3: Compute the double integral 1313xexydxdy.

 

 

 

Definition 

The average value of a function of two variables over a region R is 

favg=1areaRRf(x,y)dA.

 

Example 4: The weather map below shows an unusually moist storm system associated with the remnants of Hurricane Karl, which dumped 4-8 inches (100-200 mm) of rain in some parts of the Midwest on September 22-23, 2010. The area of rainfall measured 300 miles east to west and 250 miles north to south. Estimate the average rainfall over the entire area in those two days.

 

 

 

 

 

 

 

Exercise 4: A contour map is shown for a function f(x,y) on the rectangle R=[3,6]×[1,4].

a. Use the midpoint rule with m=3 and n=2 to estimate the value of Rf(x,y)dA.. 

b. Estimate the average value of the function f(x,y).

 

 

 

 

 

 

Example 5: Compute the double integral 0212xexydxdy.

 

 

 

 

Exercise 5: Compute the double integral 1202yeyxdxdy. 

 

 

 

Example 6: Find the volume of the solid enclosed by the surfaces, z=2x2+xy2 and the planes, z=0, y=±2, x=0 and x=2.

 

 

 

Example 7: Find the volume of the solid in the first octant bounded by z=9y2 and the plane x=5.

 

 

 

Group work:

1. Find the volume of the solid enclosed by the surfaces, z=x+x2y and the planes, z=0, x=±1, y=0 and y=2.

 

2. Find the volume of the solid in the first octant bounded by z=16x2 and the plane y=5.

 

3. Compute the double integral 0312xx2+y2dxdy.

 

4. Compute the double integral 1213yx2+y2dxdy.

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Multivariable Calculus Copyright © by Kuei-Nuan Lin. All Rights Reserved.