Prelim1-3.Ex1-6%20in%2016.1

# Prelim1-3.Ex1-6%20in%2016.1 - MULT IPLE 16 INTEGRATION 16.1...

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16 MULTIPLE INTEGRATION 16.1 Integration in Several Variables (ET Section 15.1) Preliminary Questions 1. In the Riemann sum S 8 , 4 for a double integral over R = [ 1 , 5 ] × [ 2 , 10 ] , what is the area of each subrectangle and how many subrectangles are there? SOLUTION Each subrectangle has sides of length x = 5 1 8 = 1 2 , y = 10 2 4 = 2 Therefore the area of each subrectangle is A = x y = 1 2 · 2 = 1, and the number of subrectangles is 8 · 4 = 32. 2. Estimate the double integral of a continuous function f over the small rectangle R = [ 0 . 9 , 1 . 1 ] × [ 1 . 9 , 2 . 1 ] if f ( 1 , 2 ) = 4. SOLUTION Since we are given the value of f at one point in R only, we can only use the approximation S 11 for the integral of f over R . For S 11 we have one rectangle with sides x = 1 . 1 0 . 9 = 0 . 2 , y = 2 . 1 1 . 9 = 0 . 2 Hence, the area of the rectangle is A = x y = 0 . 2 · 0 . 2 = 0 . 04. We obtain the following approximation: R f d A S 1 , 1 = f ( 1 , 2 ) A = 4 · 0 . 04 = 0 . 16 3. What is the integral of the constant function f ( x , y ) = 5 over the rectangle [− 2 , 3 ] × [ 2 , 4 ] ? SOLUTION The integral of f over the unit square R = [− 2 , 3 ] × [ 2 , 4 ] is the volume of the box of base R and height 5. That is, R 5 d A = 5 · Area ( R ) = 5 · 5 · 2 = 50 4. What is the interpretation of R f ( x , y ) d A if f ( x , y ) takes on both positive and negative values on R ? SOLUTION The double integral R f ( x , y ) d A is the signed volume between the graph z = f ( x , y ) for ( x , y ) R , and the xy -plane. The region below the xy -plane is treated as negative volume. 5. Which of (a) or (b) is equal to 2 1 5 4 f ( x , y ) dy dx ? (a) 2 1 5 4 f ( x , y ) dx dy (b) 5 4 2 1 f ( x , y ) dx dy SOLUTION The integral 2 1 5 4 f ( x , y ) dy dx is written with dy preceding dx , therefore the integration is first with respect to y over the interval 4 y 5, and then with respect to x over the interval 1 x 2. By Fubini’s Theorem, we may replace the order of integration over the corresponding intervals. Therefore the given integral is equal to (b) rather than to (a).

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