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# 6.3 Volumes of Revolution Mon March 10 PowerPoint PPT Presentation

6.3 Volumes of Revolution Mon March 10. 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. Solid of revolution.

6.3 Volumes of Revolution Mon March 10

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### 6.3 Volumes of RevolutionMon March 10

• 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

### Solid of revolution

• 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

### Disk Method

• 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

### Ex

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

### Washer Method

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

### Ex

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

### Revolving about any horizontal line

• 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)

### Ex

• 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

### Revolving about a vertical line

• 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!

### Ex

• 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

### Closure

• 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

### 6.3 Solids of RevolutionTues March 11

• 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

### Solids of Revolution

• 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

### Closure

• 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

### 6.3 Solids of Revolution ReviewWed March 12

• 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

### 6.1-6.3 Review _ questionsGraphing Calculator = Set up integral

• 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

### Closure

• HW: Ch 6 AP Questions MC #1-6 8-14 17 18 20 FRQ #1 2

• 6.1-6.3 Quiz Fri

• 2)D14) E

• 4)C18) A

• 6)C20) B

• 8)D2a)

• 10) Cb)

• 12) Cc)

### 6.1-6.3 Review

• 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

### 6.1-6.3 Review 6 questionsGraphing Calculator = Set up integral

• 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

### Closure

• Which application of the integral do you imagine would be the most useful in real world applications? Why?

• 6.1-6.3 Quiz tomorrow!