4. Multiple Integrals

4. Multiple Integrals - Math2011, Multiple Integrals 4.1...

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Unformatted text preview: Math2011, Multiple Integrals 4.1 Double Integrals Definition 4.1 Let f be a function in two variables, R be a region in the plane. The integral of f over R , denoted by R R f or R R f ( x, y ) dxdy , is defined to be the limit of X i f ( a point in the i th rectangle ) · ( area of the i th rectangle ) as the area of the rectangles approach zero. Here the region R is partitioned into rectangles which lie entirely in R . Remark 4.2 If f ( x, y ) > for all x and y , let S be the solid consisting of all points ( x, y, z ) such that ( x, y ) belongs to R and ≤ z ≤ f ( x, y ) . Then, R R f is the volume of S . Remark 4.3 Consider a certain piece of material occupying the region R . f ( x, y ) = its density (mass per unit area) at the point ( x, y ) . Then, R R f is the total mass of that piece of material. Theorem 4.4 If f is a function in two variables, R is the rectangle consisting of all points ( x, y ) such that a ≤ x ≤ b and c ≤ y ≤ d . Then, Z R f = Z d c [ Z b a f ( x, y ) dx ] dy = Z b a [ Z d c f ( x, y ) dy ] dx. proof: omitted. Example 4.5 Let f ( x, y ) = 1 + xy and R be the rectangle consisting of all points ( x, y ) such that ≤ x ≤ 1 and 1 ≤ y ≤ 2 . Evaluate R R f . solution: R R f = R 2 1 [ R 1 (1 + xy ) dx ] dy = R 2 1 (1 + y/ 2) dy = 7 / 4 . Theorem 4.6 Let f be a function in two variables. 1 1. If R is the region consisting of all points ( x, y ) such that g ( y ) ≤ x ≤ h ( y ) and a ≤ y ≤ b . Then, R R f = R b a R h ( y ) g ( y ) f ( x, y ) dxdy . 2. If R is the region consisting of all points ( x, y ) such that g ( x ) ≤ y ≤ h ( x ) and a ≤ x ≤ b . Then, R R f = R b a R h ( x ) g ( x ) f ( x, y ) dydx . proof of 1): Let c , d be numbers such that c < g ( y ) and d > h ( y ) for all a ≤ y ≤ b . Let R be the rectangle consisting of all points ( x, y ) such that c ≤ x ≤ d and a ≤ y ≤ b . Define F to be a function so that F ( x, y ) = f ( x, y ) for ( x, y ) in R and F ( x, y ) = 0 for ( x, y ) not in R . Then, R R f = R R F = R b a R d c F ( x, y ) dxdy = R b a [ R g ( y ) c F ( x, y ) dx + R h ( y ) g ( y ) F ( x, y ) dx + R d h ( y ) F ( x, y ) dx ] dy = R b a R h ( y ) g ( y ) f ( x, y ) dxdy. Example 4.7 Let f ( x, y ) = 1 + xy and R be the triangle with vertices (0 , 0) , (1 , 0) , (0 , 2) . Evaluate R R f . solution: R consists of all points ( x, y ) such that 0 ≤ x ≤ 1- y/ 2 and 0 ≤ y ≤ 2. So, R R f = R 2 R 1- y/ 2 (1 + xy ) dxdy = R 2 (1- y/ 2) + ( 1 2 (1- y/ 2) 2 y ) dy = R 2 1- y 2 / 2 + y 3 / 8 dy = 7 / 6 . Example 4.8 Let R be the region bounded by the line y =- x , the line y = 1 and the curve y = x 2 . Evaluate Z R ( x + y ) 2 dxdy. solution: R is the region consisting of all points ( x, y ) such that- y ≤ x ≤ √ y and ≤ y ≤ 1. So, R R ( x + y ) 2 dxdy = R 1 R √ y- y ( x + y ) 2 dxdy = 1 3 R 1 ( √ y + y ) 3 dy = 1 3 R 1 y 3 / 2 + 3 y 2 + 3 y 5 / 2 + y 3 dy = 351 / 560 ....
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4. Multiple Integrals - Math2011, Multiple Integrals 4.1...

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