- 247 Views
- Uploaded on
- Presentation posted in: General

6.3 Volumes of Revolution Mon March 10

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

- Do Now
- The volume of the solid whose base is the region enclosed by y = x^2 and y = 3, and whose cross sections perpendicular to the y-axis are rectangles of height y^3

- A solid of revolution is a solid obtained by rotating a region in the plan about an axis
- Pic:
- The cross section of these solids are circles

- If f(x) is continuous and f(x) >= 0 on [a,b] then the solid obtained by rotating the region under the graph about the x-axis has volume

- Calculate the volume V of the solid obtained by rotating the region under y = x^2 about the x-axis for [0,2]

- If the region rotated is between 2 curves, where f(x) >= g(x) >= 0, then

- Find the volume V obtained by revolving the region between y = x^2 + 4 and y = 2 about the x-axis for [1,3]

- When revolving about a horizontal line that isn’t y = 0, you have to consider the distance from the curve to the line.
- Ex: if you were revolving y = x^2 about y = -1, then the radius would be (x^2 + 1)

- Find the volume V of the solid obtained by rotating the region between the graphs of
f(x) = x^2 + 2 and g(x) = 4 – x^2 about the line y = -3

- If you revolve about a vertical line, everything needs to be in terms of y!
- Y – bounds
- Curves in terms of x = f(y)
- There is no choice between x or y when it comes to volume!

- Find the volume of the solid obtained by rotating the region under the graph of
f(x) = 9 – x^2 for [0,3] about the line x = -2

- Find the volume obtained by rotating the graphs of f(x) = 9 – x^2 and y = 12 for [0,3] about the line y = 15
- HW: p.381 #1-53 EOO

- Do Now
- Find the volume of the solid obtained by rotating the region between y = 1/x^2 and the x – axis over [1,4] about the x-axis

- Disk Method: no gaps
- Washer Method: gaps
- Outer – Inner
- Radii depend on the axis of revolution
- In terms of x or y depends on horizontal or vertical lines of revolution

- Find the volume of the solid obtained by rotating the region enclosed by y = 32 – 2x, y = 2 + 4x, and x = 0, about the y - axis
- HW: p.381 #1-53 AOO
- 6.1-6.3 Quiz on Friday (I won’t be in class Thurs, but will be available at break, 8th period, or after school

- Do Now
- Find the volume of the solid obtained by rotating the region between y = x^2 and y = 2x + 3 about the x-axis

- 6.1 Area between curves
- In terms of x or y
- Bounds - intersections

- 6.2 Volume using cross sections / Average Value
- V = Integral of area of cross sections
- AV = Integral divided by length of interval

- 6.3 Solids of Revolution
- With respect to different lines
- Disks vs Washers

- HW: Ch 6 AP Questions MC #1-6 8-14 17 18 20 FRQ #1 2
- Answers on powerpoint
- 6.1-6.3 Quiz Fri

- AP Answers (even):
- 2)D14) E
- 4)C18) A
- 6)C20) B
- 8)D2a)
- 10) Cb)
- 12) Cc)

- Ch 6 AP Worksheet
- 1) D6c) 5.470
- 2) C6d) 0.029
- 3) E7a) 1382.954 hours
- 4) B7b) increasing s’(100) = .029
- 5) D7c) 13.094 hours/day
- 6a) .3077d) 165th day
- 6b) 1.119

- 6.1 Area between curves
- In terms of x or y
- Bounds - intersections

- 6.2 Volume using cross sections / Average Value
- V = Integral of area of cross sections
- AV = Integral divided by length of interval

- 6.3 Solids of Revolution
- With respect to different lines
- Disks vs Washers

- Which application of the integral do you imagine would be the most useful in real world applications? Why?
- 6.1-6.3 Quiz tomorrow!