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

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

2 Course Website web1120/index.html Dan Barbasch () M1120 Class 5 September 4, / 16

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

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

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

6 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, / 16

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

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

9 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, / 16

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

11 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, / 16

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

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

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

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

16 Exercise 2 Some integrals Very useful formula: sec 2 x dx = tan x + C, sec x tan x dx = sec x + C, tan 2 x dx = (sec 2 x 1 ) dx. tan 2 x = sin2 x cos 2 x = 1 cos2 x cos 2 x = 1 cos 2 x 1 = sec2 x 1. Dan Barbasch () M1120 Class 5 September 4, / 16

17 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, / 16

18 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: Volume = b a π ( r(x) 2) dx = b a π (f (x)) 2 dx. Dan Barbasch () M1120 Class 5 September 4, / 16

19 Washers The method of washers is a (simple) variant of the method of disks. V = b a π [ r 2 (x) 2 r 1 (x) 2] dx. Exercise 1: Compute the volume of the region bounded by x = 3y 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, / 16

20 Hints to Exercises The graph of the region is The formula for the volume is V = b a πf (x) 2 dx = 2 0 π[3y/2] 2 dy. 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. V = b a π [ r 2 (x) 2 r 1 (x) 2] dx = 2 0 [ π 5 2 (5 3y/2) 2] dy. Dan Barbasch () M1120 Class 5 September 4, / 16

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

22 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: 2πr(x)h(x) x. Dan Barbasch () M1120 Class 5 September 4, / 16

23 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 V = π(r + r) 2 h πr 2 h =π [ (r 2 + 2r rh + ( r) 2 ) r 2] = The general formula is V = b a =2πr rh + π( r) 2 h 2πrh r. 2πr(x)h(x) dx. 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, / 16

24 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: Volume = b a 2πxf (x) dx. Dan Barbasch () M1120 Class 5 September 4, / 16

25 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 0 π(f (x)) 2 dx 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 0 2πxf (x) dx. Dan Barbasch () M1120 Class 5 September 4, / 16

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

27 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 = x2 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, / 16

28 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: V = πr 2 y. So V = We integrate in y. The volume element is 5 0 πx 2 dy = π 5 0 2y dy = 25π. Method of Shells: We integrate in x. The general formula is V = 2πrh x, and the volume is 10 V = 2πx (5 x 2 ) dx = 2π (5 x x 4 ) 10 = 25π. 8 0 Check the calculations, I skipped some arithmetic! Dan Barbasch () M1120 Class 5 September 4, / 16

29 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 V (h) = h 0 π(2y) dy = πh2. So dv dt = 2πh dh dt, dh dt = 1 dv 2πh dt, and we can evaluate dh dt = 3 8π in/sec. Dan Barbasch () M1120 Class 5 September 4, / 16

30 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 9x rotating the region bounded by y =, x = 0, x = 3 about the x 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, / 16

31 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, / 16

32 Volume of a Right Circular Cone Exercise: Compute the volume of a right circular cone of radius r and height h. Solution: Step 1: Find a region which when rotated about and axis gives the desired cone. Answer: Rotate the region in the first quadrant bounded above by y = h, below by x = r hy. The picture is the same as for exercise 1. Step 2: Disks: V = Shells: V = h 0 r 0 ( r ) 2 π h y r 2 y 3 dy = π h 2 3 2πx (h hr ) x h 0 = πr 2 h 3. dx = 2π (h x 2 2 h r x 3 ) r = πr 2 h Dan Barbasch () M1120 Class 5 September 4, / 16

33 Volume of a Right Circular Cone Slices: The circular cone of radius r and height h can be written as equation z 2 ( = h2 r x 2 + y 2) with 0 z h. Slice it perpendicular to the 2 x axis. The cross sections are hyperbolas; at x = a, you get z 2 = h2 y 2 + h2 a 2. The volume is then r 2 r 2 V = r 0 A(x) dx. The area bounded by such a hyperbola is challenging to compute. Dan Barbasch () M1120 Class 5 September 4, / 16