M1120 Class 5 Dan Barbasch September 4, 2011 Dan Barbasch () - PowerPoint PPT Presentation

M1120 Class 5 Dan Barbasch September 4, 2011 Dan Barbasch () M1120 Class 5 September 4, 2011 1 / 16

Course Website http://www.math.cornell.edu/ ˜ web1120/index.html Dan Barbasch () M1120 Class 5 September 4, 2011 2 / 16

Method of Slices A special case of what is called Cavalieri’s Principle. (graphics courtesy of Allen Back) � b V = A ( x ) dx a Start with a region (e.g. a disk) of Area A Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices Thicken it up vertically a distance ∆ y . Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices The volume is A ∆ y . Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices This even works if you thicken up at a slant as long as the (vertical) height is ∆ y . Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices This is the basis of both Cavalieri and the method of disks. Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices Start with a line where x measures distance along the line. Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices Start with a line where x measures distance along the line. Keep track of the (perpendicular to the line) cross sectional area A ( x ). Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Method of Slices � b Volume(Body) = A ( x ) dx . a Dan Barbasch () M1120 Class 5 September 4, 2011 3 / 16

Problems using Slices Exercise 1: [(12) in the text] Find the volume of a pyramid with a square base of area 9 and height 5 . Exercise 2: [(6a) in the text] Find the volume of the solid which lies between planes perpendicular to the x -axis between x = ± π 3 , and cross sections (perpendicular to the x − axis) circular disks with diameter running from the curve y = tan x to y = sec x . Dan Barbasch () M1120 Class 5 September 4, 2011 4 / 16

Exercise 2 y = tan x Dan Barbasch () M1120 Class 5 September 4, 2011 5 / 16

Exercise 2 y = tan x y = sec x Dan Barbasch () M1120 Class 5 September 4, 2011 5 / 16

Exercise 2 y = tan x y = sec x y = sec x and y = tan x together. Dan Barbasch () M1120 Class 5 September 4, 2011 5 / 16

Exercise 2 The diameter at x is d ( x ) = sec x − tan x . The area of the cross section at x is A ( x ) = π r ( x ) 2 = π ( d ( x ) / 2) 2 = π (sec x − tan x ) 2 . 4 The volume is � π/ 3 π 4 (sec x − tan x ) 2 dx . V = − π/ 3 √ � � 6 3 − π π The value of the integral is . 6 Dan Barbasch () M1120 Class 5 September 4, 2011 5 / 16

Exercise 2 Some integrals � sec 2 x dx = tan x + C , � sec x · tan x dx = sec x + C , � � � tan 2 x dx = sec 2 x − 1 � dx . Very useful formula: cos 2 x = 1 − cos 2 x tan 2 x = sin 2 x 1 cos 2 x − 1 = sec 2 x − 1 . = cos 2 x Dan Barbasch () M1120 Class 5 September 4, 2011 5 / 16

Disks and Washers Rotate the area in the picture on the left about the x − axis. The volume of the resulting body on the right is computed by the method of disks: Dan Barbasch () M1120 Class 5 September 4, 2011 6 / 16

Disks and Washers Rotate the area in the picture on the left about the x − axis. The volume of the resulting body on the right is computed by the method of disks: � b � b π ( f ( x )) 2 dx . r ( x ) 2 � � Volume = π dx = a a Dan Barbasch () M1120 Class 5 September 4, 2011 6 / 16

Washers The method of washers is a (simple) variant of the method of disks. � b r 2 ( x ) 2 − r 1 ( x ) 2 � � V = π dx . a Exercise 1: Compute the volume of the region bounded by x = 3 y 2 , x = 0 , and y = 2 revolved about the y − axis. Exercise 2: Write a definite integral which computes the volume of the solid obtained by revolving the region in the previous exercise about the axis x = 5 . Dan Barbasch () M1120 Class 5 September 4, 2011 7 / 16

Hints to Exercises The graph of the region is � b � 2 π f ( x ) 2 dx = π [3 y / 2] 2 dy . The formula for the volume is V = a 0 The formula in blue is the general formula ; but applied to this problem, integration is in y not x ; so we adjust accordingly. For the second problem the region is the same, and the general formula is “the same” too. Question: What changes? Answer: In this case we are dealing with washers, and the radius must be adjusted according to the axis of rotation. � 2 � b r 2 ( x ) 2 − r 1 ( x ) 2 � � 5 2 − (5 − 3 y / 2) 2 � � V = π dx = π dy . 0 a Dan Barbasch () M1120 Class 5 September 4, 2011 8 / 16

Question: What if we want to integrate in x ? Dan Barbasch () M1120 Class 5 September 4, 2011 9 / 16

Method of Shells This is the heart of the method of shells. Here is a picture Question: What is the volume of the body between two concentric cylinders of height h , and respective radii r + ∆ r and r ? Answer: 2 π r ( x ) h ( x )∆ x . Dan Barbasch () M1120 Class 5 September 4, 2011 10 / 16

Method of Shells The formula is “obtained by opening up the shell:” ∆ V ≈ length × height × width A somewhat more rigorous argument is the following calculation: The volume of the region between two concentric cylinders of height h , and respective radii r + ∆ r and r is ( r 2 + 2 r ∆ rh + (∆ r ) 2 ) − r 2 � ∆ V = π ( r + ∆ r ) 2 h − π r 2 h = π � = =2 π r ∆ rh + π (∆ r ) 2 h ≈ 2 π rh ∆ r . The general formula is � b V = 2 π r ( x ) h ( x ) dx . a I use blue again to emphasize that this is the “general formula” . You have to adjust according to the problem. Dan Barbasch () M1120 Class 5 September 4, 2011 10 / 16

Method of Shells For the body of revolution obtained by rotating the region in the picture on the left about the y -axis, the method of shells gives: � b Volume = 2 π xf ( x ) dx . a Dan Barbasch () M1120 Class 5 September 4, 2011 10 / 16

Summary For the region between x = a , x = b , bounded above by y = f ( x ) ≥ 0 and below by the x − axis, the volume obtained by revolving about the x -axis is given by the method of disks � b π ( f ( x )) 2 dx 0 The volume of the body obtained by revolving the region before with a ≥ 0 , b ≥ 0 rotated about the y − axis is given by the method of shells: � b 2 π xf ( x ) dx . 0 Dan Barbasch () M1120 Class 5 September 4, 2011 11 / 16

Solution to Exercises 1,2 Method of Shells � b The general formula is a 2 π h ( x ) r ( x ) dx . For the first problem, h ( x ) = 5 − y = 5 − 2 x / 3 , r ( x ) = x : � 3 � 5 − 2 x � V = 2 π x dx = 6 π. 3 0 For the second problem, h ( x ) = 5 − y = 5 − 2 x / 3 as before, but the radius is r ( x ) = 5 − x . The volume is � 3 V = 2 π (5 − x )(5 − 2 x / 3) dx = 24 π. 0 Dan Barbasch () M1120 Class 5 September 4, 2011 12 / 16

Problems from section 6.1 Exercise 1: [(22) in the text] Find the volume of the body obtained by rotating the region bounded by y = 2 √ x , x = 0, y = 2, about the x − axis. Exercise 2: [(40) in the text] Find the volume of the body obtained by rotating the region bounded by y = 2 √ x , x = 0, y = 2, is revolved about the x -axis. Exercise 3: [(56) in the text] Find the volume of the bowl which has a shape generated by revolving the graph of y = x 2 2 between y = 0 and y = 5 about the y -axis. Water is running in into the bowl at 3 cubic units per second. How fast will the water be rising when it is 4 units deep? Dan Barbasch () M1120 Class 5 September 4, 2011 13 / 16

Solution to Exercise 3: The bowl is obtained by rotating the region in the first quadrant between the y − axis and y = x 2 / 2. Method of Disks: We integrate in y . The volume element is � 5 � 5 ∆ V = π r 2 ∆ y . So V = π x 2 dy = π 2 y dy = 25 π . 0 0 Method of Shells: We integrate in x . The general formula is ∆ V = 2 π rh ∆ x , and the volume is √ � √ 10 10 5 − x 2 5 x 2 2 − x 4 � � � �� � V = 2 π x dx = 2 π = 25 π. � 2 8 � 0 0 Check the calculations, I skipped some arithmetic! Dan Barbasch () M1120 Class 5 September 4, 2011 14 / 16

For the second part of the problem, we need the volume V as a function of the height h . We know that dV dt = 3 . We can use h = y ; FTC implies � h 0 π (2 y ) dy = π h 2 . So V ( h ) = dV dt = 2 π hdh dh 1 dV dt , dt = dt , 2 π h dt = 3 and we can evaluate dh 8 π in / sec . Dan Barbasch () M1120 Class 5 September 4, 2011 14 / 16

Problems from Section 6.2 Use the method of shells in the following exercises. Exercise 1: [(6) in the text] Compute the volume of the body generated by 9 x rotating the region bounded by y = √ , x = 0, x = 3 about the x 3 + 9 y -axis. Exercise 2: [(10) in the text] Compute the volume of the body generated by rotating the region bounded by y = 2 − x 2 , x = 0, y = x 2 about the y -axis. Exercise 3: [(22) in the text] Compute the volume of the body generated by rotating the region bounded by y = √ x , y = 0, y = 2 − x about the x -axis. Dan Barbasch () M1120 Class 5 September 4, 2011 15 / 16

Problems from Section 6.2 Use the method of shells in the following exercises. Exercise 4: [(24) in the text] Consider the region bounded by y = x 3 , y = 8, x = 0 . Compute the volume of the body generated by this region when rotated about the 1 y -axis. 2 line x = 3 . 3 line x = − 2 . 4 x -axis. 5 line y = 8 . 6 line y = − 1 . Dan Barbasch () M1120 Class 5 September 4, 2011 15 / 16