Volume: The Disk Method Using the integral to find volume.
If a region in a plane is revolved about a line, the resulting solid is a solid of revolution and the line is called the axis of revolution. y = x
Volume of a Solid of Revolution In order to find the volume of a solid of revolution, you must first look at the cross sections of the solid. y = x The Cross Section is a circle so: A = πr 2 What is the r? A x = π x 2
Volume of a Solid of Revolution Then after finding a formula for the area of the cross section, you will integrate this function to find the volume. y = x A x = π x 2 A x = πx b A(x) a dx 10 πx 0 dx
Simplifying the formula. Whenever we revolve the area around an axis of revolution, our cross sections are circles. Since our radius will be a function of x, we can rewrite R as R(x). V = a b V = π R x 2 dx a b A(x)dx b V = π R x 2 dx a A Region to revolve. B Y = R(x) Note that R(x) represents the radius of the revolved cross section.
Example 1: Revolve the area of y = 4 x 2 from x = 2 to x = 2 around the x axis. b V = π R x 2 dx a y = 4 x 2 R x = 4 x 2 V = π 2 2 4 x 2 2 dx
Example 2: Find the volume formed by revolving the region f x = sin x about the x-axis from 0 to π.
The solid formed from f x = sin x
What is the volume of the solid formed by revolving about the y-axis from y = 0 to y = 4? Step One: determine the radius of a cross section. y = 16 x 2 y 2 = 16 x 2 x = 16 y 2 Step Two: Set up same formula and integrate with respect to y.
Homework Section 7.2 (1-4 and 7-10)
8.
9.
Day 2: Revolving about a line that is not the x or y axis. If you are revolving around a line that is parallel to the x axis, you will integrate with respect to x. If you are revolving around a line that is parallel to the y axis, you will integrate with respect to y.
What is your radius? When doing a volume problem, you need to ask What is the radius of the circle I am creating? When revolving about the Line y = 2.. F(x) R(1) = R x = y = 2 V = π F x 2 2 dx 3 3
Revolving about a line that is not the x-axis. Example: Find the volume of the solid formed by revolving the region bounded by f x = 2 x 2 and g x = 1 about the line y = 1. To determine the radius.. R x = f x g x = 2 x 2 1 = 1 x 2 If you let f x = g x, you can see that these two lines intersect at 1 and 1.
Revolving around Vertical line (not y-axis) What is the volume of the solid formed by revolving the lines x = 5 and x = y 2 + 14 around the line x = 5? 4 3 2 1 R y = y 2 + 14 5 R y = y 2 + 9-1 -2-3 -4 5 10 15 V = π 3 3 y 2 + 9 2 dy
Homework Section 7.2 (6 (revolve around y=2), 11A, 11C, 12B, 12D, 16, 22,)
Revolve the region around Y = 2. 4 x2 4 = 2 x = ±2 2
V = π 2 2 2 2 4 x2 4 2 2 dx V = π 2 2 2 2 x2 4 + 2 2 dx V = 128 2 15 π
Day 3: The Washer Method So far, all regions that we revolve have always been in contact with the access of revolution. What if we revolve a region that is not always in contact with the access of revolution?
The Washer Method The washer method is used to revolve a plane around an axis that it is not touching (either partly or entirely).
Imagine for a second that the rectangle touched the axis of revolution and you formed a solid cylinder as before. The volume would be π R 2 dx Now consider the area that is cut out from the cylinder. The radius of this empty cylinder is r. The volume of the smaller solid will be π r 2 dx If you subtract the smaller cylinder from the larger, you will have the volume. π R 2 dx π r 2 dx = π R 2 r 2 dx
V = π R x 2 r x 2 dx b a
Main Things to Consider in this type of Problem 1. IMAGINE YOU ARE REVOLVING A SOLID AREA THAT ALWAYS TOUCHES AXIS IN ORDER TO FIND R(X). 2. THINK ABOUT THE SMALLER AREA THAT IS CUT OUT IN ORDER TO FIND THE r(x). V = π R x 2 r x 2 dx b a
Set up this integral.. Revolve the area around the x-axis. V = π R x 2 r x 2 dx b a
Example 3 (Washer Method) Find the volume of the solid formed by revolving the region bounded by the graphs of y = x and y = x 2 about the x- axis.
y = x and x=4 revolved around x=7. x = y 2 R y = 7 y 2 6 r y = 7 4 = 3 2 π 7 y 2 2 3 2 0 dy
y = 2x 2 and x = 2 and x axis around y = 9. R x = 9 r x = 9 2x 2 2 π 9 2 9 2x 2 2 0 dx
Revolve the area enclosed by the above Graphs around the line y = 8. R x = 8 (x 2 + 1) r x = 8 ( x 2 + 2x + 5) 2 π 7 x 2 2 1 x 2 2x + 3 2 dx
What if we tried to revolve around the line x = 8??? What would be the problem we would encounter?
Questions to ask yourself when solving. 1. Is the axis of revolution parallel to X axis dx or Y axis dy? 2. What is your solid/big radius? R X 3. What is the radius of the portion you will cut out? r(x)
Homework Section 7.2 (11d, 12c, 13a, 13b, 14a, 14b, 17)
11d.
13A. Revolve around X-axis
13b.
14A. Revolve around the X-Axis
14B.
17. y = 1 1+x, y = 0, x = 0, x = 3 Revolve around y = 4
After discussing these homework problems give students the following AP problems. 2001 AB Free Response #1 2002 AB Free Response #1 *Use Calculus in Motion to explain solutions.*
Summary Revolving around the x axis or y axis. Region always touches axis. Revolving around an axis parallel to x axis or y axis. Region always touches axis. Revolving around an axis that is not always in contact with region. Washer Method.
Area and Volume Review Set up the equations for revolving around the following lines? Y = 4 π 2 4 x 2 2 dx 2 X-Axis π 2 4 2 x 2 2 dx 2 Y = 6 π 2 6 x 2 2 6 4 2 dx 2
Area and Volume Review In words, tell me the difference between revolving around the following lines? A. Y = 4 B. X Axis C. Y = 6
Cross Sections and Volume In the following picture the axis of revolution is parallel to what? While the cross sections are perpendicular to what? If the cross sections are perpendicular to the x-axis... Volume = 0 1 A x dx **Note A(x) is a circle. Thus ** A x = π r x 2
In the following picture, The axis of revolution is parallel to what? The cross sections are perpendicular to what? If the cross sections are perpendicular to the y-axis Volume = 0 8 A y dy **Note A(y) is a circle. Thus ** A y = π r y 2
How would we find the volume of the following? Volume = 0 h A x dx Volume = h A y dy 0
Example: Find the volume of the solid whose base is bounded by the lines: f x = 1 x, g x = 1 + x and x = 0. 2 2 (The cross sections are perpendicular to the x-axis and form equilateral triangles). Area = 4 3 2 base
The base will equal the difference in the Y values.
Homework *Write/Find better cross section homework for next year.* Section 7.2 (61, 62 A-C, 67 73 odd) (Set up Equations ONLY!!) 2000 AP Test 1C 2003 AP (1) (Use Calculus in Motion to explain solutions)
After discussing 2000 AP Test 1C, 2003 AP (1). Do the following AP Problems AB 2007 (1) AB 2009 (4)