Introduction to Multiple Integrals Chapters 5.15.2 and parts of - PowerPoint PPT Presentation

Introduction to Multiple Integrals Chapters 5.1–5.2 and parts of 5.3–5.5 Prof. Tesler Math 20C Fall 2018 Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 1 / 26

Indefinite integrals with multiple variables Consider � e ax dx = e ax a + C In the input, dx says x is the integration variable. a is constant. In the result, C is a constant (does not depend on x ). Applying d / dx to the result gives back the integrand: � e ax � d = e ax a + C dx Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 2 / 26

Indefinite integrals with multiple variables Let x , y , z be variables, and consider � ( 2 xy + z ) dz = 2 xyz + z 2 2 + C ( x , y ) In the input: dz says z is the integration variable. x , y are treated as constants while doing the integral. In the result: The integration “constant” does not depend on the integration variable z , but it might depend on the other variables x , y ! So it’s a function, C ( x , y ) . Applying ∂/∂ z to the result gives back the integrand: 2 xyz + z 2 � � ∂ 2 + C ( x , y ) = 2 xy + z ∂ z Note ∂ ∂ z C ( x , y ) = 0 for all functions of x and y . Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 3 / 26

Definite integrals with multiple variables � b z = b 2 xyz + z 2 �� � � ( 2 xy + z ) dz = � 2 � a z = a As a definite integral: The limits a , b may depend on the other variables, x and y . Specify limits as z = a and z = b instead of just a and b : b 2 xyz + z 2 �� � � Don’t do this: � 2 � a This is ambiguous; it doesn’t say which of x , y , or z is the variable to set equal to a and to b . No need for the integration constant; it will cancel upon subtracting the antiderivatives at the two limits. Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 4 / 26

Definite integrals with multiple variables Method 1: Antiderivative at upper limit minus at lower limit � x + y z = x + y 2 xyz + z 2 �� � � ( 2 xy + z ) dz = � 2 � 0 z = 0 2 xy ( x + y ) + ( x + y ) 2 2 xy ( 0 ) + 0 2 � � � � = − 2 2 = 2 xy ( x + y ) + ( x + y ) 2 2 Method 2: Subtract term-by-term � x + y z = x + y 2 xyz + z 2 �� � � ( 2 xy + z ) dz = � 2 � 0 z = 0 = 2 xy (( x + y ) − 0 ) + ( x + y ) 2 − 0 2 2 = 2 xy ( x + y ) + ( x + y ) 2 2 Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 5 / 26

Iterated integrals This is a triple integral : � 1 � 2 x � x + y 2 xy dz dy dx 0 0 x Group it like this, with parentheses: � 1 � � 2 x � � x + y � � 2 xy dz dy dx 0 x 0 � Match up integral signs and differentials (like dx ) inside-to-out, not left-to-right: Inside integral: z goes from 0 to x + y Middle integral: y goes from x to 2 x Outside integral: x goes from 0 to 1 . The limits for each variable can only depend on variables that are farther outside than they are: � b � g 2 ( x ) � h 2 ( x , y ) f ( x , y , z ) dz dy dx g 1 ( x ) h 1 ( x , y ) a Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 6 / 26

Iterated integrals � 1 � 2 x � x + y I = 2 xy dz dy dx 0 0 x Evaluate the inside integral: � x + y 2 xy dz = ( 2 xyz ) | z = x + y = 2 xy (( x + y ) − 0 ) = 2 xy ( x + y ) z = 0 0 Replace the inside integral by what it evaluates to: � 1 � 2 x I = 2 xy ( x + y ) dy dx 0 x Now it’s a double integral. Iterate! There’s a new inside integral; repeat this until all integrals are evaluated. Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 7 / 26

Iterated integrals � 1 � 2 x � x + y � 1 � 2 x I = 2 xy dz dy dx = 2 xy ( x + y ) dy dx 0 x 0 0 x Iterate! The new inside integral is: � 2 x � 2 x ( 2 x 2 y + 2 xy 2 ) dy 2 xy ( x + y ) dy = x x y = 2 x x 2 y 2 + 2 xy 3 �� � � = � 3 � y = x = x 2 (( 2 x ) 2 − x 2 ) + 2 x (( 2 x ) 3 − x 3 ) 3 = x 2 ( 3 x 2 ) + 2 x ( 7 x 3 ) = 3 x 4 + 14 x 4 = 23 x 4 3 3 3 � 1 23 x 4 Replace the inside integral by its evaluation: I = 3 dx 0 Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 8 / 26

Iterated integrals � 1 � 2 x � x + y � 1 23 x 4 I = 2 xy dz dy dx = · · · = dx 3 0 x 0 0 Now it’s down to a single integral. Finally, x = 1 = 23 ( 1 5 − 0 5 ) I = 23 x 5 � = 23 � � 15 15 15 � x = 0 Going back to the original problem: � 1 � 2 x � x + y 2 xy dz dy dx = 23 15 0 x 0 Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 9 / 26

Rectangle D = [ 0 , 2 ] × [ 1 , 3 ] D = [ 0 , 2 ] × [ 1 , 3 ] is the filled-in rectangle in y the xy plane with 0 � x � 2 and 1 � y � 3 . Our book often uses R for rectangle and D x for any 2-dimensional shape. This is called the Cartesian product . In set notation: D = [ 0 , 2 ] × [ 1 , 3 ] = { ( x , y ) : 0 � x � 2 and 1 � y � 3 } = { ( x , y ) | 0 � x � 2 and 1 � y � 3 } In set notation, some books use a colon : and others use a bar | B = [ a , b ] × [ c , d ] × [ e , f ] is a filled-in box in 3D: B = { ( x , y , z ) : a � x � b , e � z � f } . c � y � d , and Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 10 / 26

Volume under z = 2 x + 1 and above rectangle D = [ 0 , 2 ] × [ 1 , 3 ] z The coordinates on the plane z = 2 x + 1 (2,1,5) (2,3,5) above the corners of the rectangle D are z=2x+1 ( x , y ) z (0,1,1) (0,3,1) ( 0 , 1 ) 1 (0,1,0) (0,3,0) ( 0 , 3 ) 1 y ( 2 , 1 ) 5 D x ( 2 , 3 ) 5 (2,1,0) (2,3,0) Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 11 / 26

Volume under z = 2 x + 1 and above rectangle D = [ 0 , 2 ] × [ 1 , 3 ] Method: split into pieces with known volumes z (2,1,5) (2,3,5) (0,1,1) (0,3,1) (0,3,0) y (2,3,1) x (2,1,0) (2,3,0) The plane z = 1 splits the volume into two parts: Bottom: box 2 · 2 · 1 = 4 Top: half a box ( 2 · 2 · 4 ) / 2 = 8 Total: 12 If x , y , z are in cm, this is 12 cm 3 . Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 12 / 26

Volume under z = 2 x + 1 and above rectangle D = [ 0 , 2 ] × [ 1 , 3 ] Method: dA = dy dx dx y dy x Split up D by making a grid with closely spaced horizontal lines and closely spaced vertical lines. dA is differential area. It can be dA = dx dy or dA = dy dx . Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 13 / 26

Volume under z = 2 x + 1 and above rectangle D = [ 0 , 2 ] × [ 1 , 3 ] Method: dA = dy dx Let D be a region in the xy plane and let f ( x , y ) � 0 on D . Volume above patch at ( x , y ) and below z = f ( x , y ) is (height)(differential area) = f ( x , y ) dA �� f ( x , y ) dA is the volume above D and below z = f ( x , y ) . D For our current example, �� ( 2 x + 1 ) dA = 12 D Use parentheses around 2 x + 1 , since it’s multiplied by dA . �� Do not write it as 2 x + 1 dA D Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 14 / 26

Volume under z = f ( x , y ) and above region D Cavalieri’s Principle z (2,1,5) (2,3,5) Let E be a 3D region. Let a � x � b be the range of x in E . (0,3,1) Slice E at many values of x ; (0,3,0) e.g., set ∆ x = ( b − a ) / n , y slice E at x = a , a + ∆ x , a + 2 ∆ x , . . . , b , x and let n → ∞ . (2,1,0) (2,3,0) The infinitesimal cross-section at x = x 0 (called an x -slice ) has area A ( x 0 ) , thickness dx , and volume A ( x 0 ) dx . The total volume of E is �� � b � b V = f ( x , y ) dA = a (area of cross-section at x ) dx = a A ( x ) dx D This can also be done with y or z cross-sections. Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 15 / 26

Volume under z = 2 x + 1 and above rectangle D = [ 0 , 2 ] × [ 1 , 3 ] Method: dA = dy dx z (2,1,5) (2,3,5) (0,3,1) (0,3,0) y x (2,1,0) (2,3,0) First, dA = dy dx . Set up the integral with x , y limits coming from D : � 2 � 3 �� ( 2 x + 1 ) dA = ( 2 x + 1 ) dy dx 0 1 D The slice at x has infinitesimal thickness dx , � 3 � 3 area 1 ( 2 x + 1 ) dy , and volume ( 1 ( 2 x + 1 ) dy ) dx . Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 16 / 26

Volume under z = 2 x + 1 and above rectangle D = [ 0 , 2 ] × [ 1 , 3 ] Method: dA = dy dx z (2,1,5) (2,3,5) � 2 � 3 �� (0,3,1) ( 2 x + 1 ) dA = ( 2 x + 1 ) dy dx (0,3,0) 0 1 D y x (2,1,0) (2,3,0) Inside: � 3 y = 3 � ( 2 x + 1 ) dy = ( 2 x + 1 ) y y = 1 = ( 2 x + 1 )( 3 − 1 ) = 2 ( 2 x + 1 ) � � 1 Outside: � 2 2 ( 2 x + 1 ) dx = 2 ( x 2 + x ) � x = 2 x = 0 = 2 (( 2 2 − 0 2 ) + ( 2 − 0 )) � 0 = 2 ( 4 + 2 ) = 12 Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 17 / 26

Volume under z = 2 x + 1 and above rectangle D = [ 0 , 2 ] × [ 1 , 3 ] Method: dA = dx dy z (2,1,5) (2,3,5) (0,3,1) (0,3,0) y x (2,1,0) (2,3,0) Next, dA = dx dy . Set up the integral with x , y limits coming from D : � 3 � 2 �� ( 2 x + 1 ) dA = ( 2 x + 1 ) dx dy 1 0 D The slice at y has infinitesimal thickness dy , � 2 � 2 area 0 ( 2 x + 1 ) dx , and volume ( 0 ( 2 x + 1 ) dx ) dy . Prof. Tesler 5.1–5.2 Multiple Integrals – Intro Math 20C / Fall 2018 18 / 26