Transcript 6.3 Volumes of Revolution Fri March 15

6.3 Volumes of Revolution
Tues March 10
• Do Now
• Find 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 yaxis are rectangles of height y^3
Solid of revolution
• A solid of revolution is a solid obtained by
rotating a region in the plane 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
V = p ò r dx = p ò f (x) dx
b 2
b
a
a
2
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
V = p ò (r
b
a
2
outer
-r
2
inner
) dx
= p ò ( f (x) - g(x) ) dx
b
a
2
2
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 Revolution
Wed 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
HW Review: p.381 #1-53
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 Mon or Tues?
6.3 Solids of Revolution Review
Thurs 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
HW Review: p.381 #1-53
6.1-6.3 Review _ questions
Graphing 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
• Answers on powerpoint
• 6.1-6.3 Quiz Tues
6.1-6.3 Review
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AP Answers (even):
2) D
4) C
6) C
8) D
10) C
12) C
14) E
18) A
20) B 2
3
2a) ò 0 (4x - x )dx
8
y
2
3
b) p ò ( y ) - ( )2 dy
0
4
c)
p ò 0 (20 - x 3 )2 - (20 - 4x)2 dx
2
6.1-6.3 Review
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Ch 6 AP Worksheet
1) D
6c) 5.470
2) C
6d) 0.029
3) E
7a) 1382.954 hours
4) B
7b) increasing s’(100) = .029
5) D
7c) 13.094 hours/day
6a) .307
7d) 165th day
6b) 1.119
6.1-6.3 Review 6 questions
Graphing 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 Tues!