4. Multiple Integrals

# 4. Multiple Integrals - Math2011 Multiple Integrals 4.1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online