# Slide_10 - MA1104 Multivariable Calculus Lecture 10 Dr KU...

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 10 Dr KU Cheng Yeaw Monday Feb 16, 2009 Recap Last lecture... • we learned how to use the Method of Lagrange Multiplier to optimize functions of several variables, say f ( x,y,z ) subject to the constraint g ( x,y,z ) = k . • To solve the optimization problem, we have to solve the equations O f ( x,y,z ) = λ O g ( x,y,z ) , g ( x,y,z ) = k. Overview In this lecture, • we introduce integrals for functions of two variables. • we first define double integral over a rectangle region. We introduce iterated integral to help us compute double integrals via Fubini’s Theorem. We then consider integral over a more general region. Double Integrals Over Rectangles Having studied derivatives for functions of several variables and their applications, we now turn to introducing the idea of integral for functions of several variables. It turns out that these ideas are useful in many practical problems. Recall that in Calculus of Single Variable, our attempt to find the area under a curve led to the definition of a definite integral. We now seek to find volume under a surface and in the process we arrive at the definition of a double integral . We start by reviewing how we arrive at the definite integral of functions of a single variable: Step 1. Suppose f ( x ) is defined for a ≤ x ≤ b . We divide the interval [ a,b ] into n subintervals of equal size 4 x = b- a n . Step 2. We choose sample points x * i from these subintervals and form the Riemann Sum n X i =1 f ( x * i ) 4 x. Step 3. Take the limit of such sum as n → ∞ to obtain the definite integral of f from a to b : Z b a f ( x ) dx = lim n →∞ n X i =1 f ( x * i ) 4 x. In the special case where f ( x ) ≥ , the integral R b a f ( x ) dx represents the area under the curve f ( x ) from a to b . Volume and Double Integral Suppose f ( x,y ) is a function of two variables defined on a closed rectangle R = [ a,b ] × [ c,d ] = { ( x,y ) ∈ R 2 : a ≤ x ≤ b,c ≤ y ≤ d } . Suppose f ( x,y ) ≥ . The graph of f is a surface with z = f ( x,y ) above R Let S be the solid that lies above R and under the graph of f . How can we find the volume of S ? We can estimate the volume of S as follows: Step 1. divide the rectangle R into subrectangles. We do this by • dividing the interval [ a,b ] into m subintervals [ x i- 1 ,x i ] of equal length 4 x = b- a m , and • dividing the interval [ c,d ] into n subintervals [ y j- 1 ,y j ] of equal length 4 y = d- c n . Form subrectangles R ij = [ x i- 1 ,x i ] × [ y j- 1 ,y j ] , for all 1 ≤ i ≤ m, 1 ≤ j ≤ n. Each of these subrectangles has area 4 A = 4 x 4 y . Step 2. Choose a sample point ( x * ij ,y * ij ) in each R ij . Then approximate the part of S lies above R ij by a thin rectangle box with base R ij and height f ( x * ij ,y * ij ) . The volume of this box is given by f ( x * ij ,y * ij ) 4 A. It follows that by adding the volumes of all these thin boxes, we get an approximation of the total volume of S : V ≈ m X i =1 n X j =1 f ( x * ij ,y * ij ) 4 A. Our intuition tells us that the approximation becomes better as m,n → ∞ . So we would expect V = lim m,n →∞ m X i =1 n X j =1 f ( x * ij ,y * ij ) 4 A. We make the following definition:...
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Slide_10 - MA1104 Multivariable Calculus Lecture 10 Dr KU...

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