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Chapter 15 Multiple Integrals
15.1 Double Integrals over Rectangles 15.2 Iterated Integrals 15.3 Double Integrals over General Regions 15.4 Double Integrals in polar coordinates 15.5* Applications of Double Integrals 15.6* Surface Area
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15.1 Double Integrals over Rectangles
Volumes and Double Integrals A function f of two variables defind on a closed rectangle and we suppose that The graph of f is a surface with equation Let S be the solid that lies above R and under the graph of f ,that is ,
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(See Figure 1)Find the volume of S
1) Partition: The first step is to divide the rectangle R into subrectangles. Each with area
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2) Approximation: A thin rectangular box: Base: Hight: We can approximate by
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3) Sum: A double Riemann sum 4) Limit:
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The sufficient condition of integrability:
Definition The double integral of f over the rectangle R is if this limit exists. The sufficient condition of integrability: Theorem1. is integral on R
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Theorem2. and f is discontinuous only on a finite number of smooth curves is integral on
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Example 1 If evaluate the integral Solution
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15.2 Iterated Integrals Partial integration with respect to y defines a function of x: We integrate A with respect to x from x=a to x=b, we get
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The integral on the right side is called an iterated integral
and is denoted by Thus Similarly
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Fubini’s theorem If f is continuous on the rectangle
then More generally, this is true that we assume that f is bounded on R , f is discontinuous only on a finite number of smooth curves, and the iterated integrals exist. The proof of Fubini’s theorem is too difficult to include In our class.
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If f (x,y) ≥0,then we can interpret the double integral
as the volume V of the solid S that lies above R and under the surface z=f(x, y). So
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Or
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Example 1 Solution 1
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Solution 2
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Example 2 Solution
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Example 3 Solution
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Specially If Then
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15.3 Double Integrals over General Regions
Suppose that D is a bounded region which can be enclosed in a rectangular region R. D D A new function F with domain R:
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If the integral of F exists over R, then we define the
double integral of f over D by
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If the integral of F exists over R, then we define the
double integral of f over D by A plane region D is said to be of type I if Where and are continuous on [a,b] Some examples of type I
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Evaluate where D is a region of type I
A new function F with domain R:
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If f is continuous on type I region D such that
then
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A plane region D is said to be of type II if
Where and are continuous on [a,b] Some examples of type II
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If f is continuous on type II region D such that
then
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Example 1 Evaluate ,where D is the region bounded by the parabolas Solution Type I
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Type II
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Properties of Double Integral Suppose that functions f and g are continuous on a bounded closed region D. Property 1 The double integral of the sum (or difference) of two functions exists and is equal to the sum (or difference) of their double integrals, that is, Property 2 Property 3 where D is divided into two regions D1 and D2 and the area of D1 ∩ D2 is 0.
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Property 4 If f(x, y) ≥0 for every (x, y) ∈D, then Property 5 If f(x, y)≤g(x, y) for every (x, y) ∈D, then Moreover, since it follows from Property 5 that hence
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Property 6 Suppose that M and m are respectively the maximum and minimum values of function f on D, then where S is the area of D. Property 7 (The Mean Value Theorem for Double Integral) If f(x, y) is continuous on D, then there exists at least a point (ξ,η) in D such that where S is the area of D. f (ξ,η) is called the average Value of f on D
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Example 2 Evaluate ,where D is the region bounded by the parabolas Solution Type II
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Type I
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Example 3 Evaluate ,where D is the region bounded by the parabolas Solution Type I
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15.4 Double Integrals in polar coordinates
A polar rectangle
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where The “center” of the polar subrectangle has polar coordinates
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The area of is
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Change of polar coordinate in a double integral
If f is continuous on a polar rectangle R given by , ,where , then
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1.If f is continuous on a polar region of the form
then
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2.If f is continuous on a polar region of the form
then
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3.If f is continuous on a polar region of the form
then
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Evaluate ,where D is the region
bounded by the circles Example 1 Solution 1 2 o
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Example 2 Evaluate where D = Solution In polar coordinates, the equation x2 + y2 –2x = 0 becomes The disc D is given by D’: Therefore
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