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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.
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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 • Answers on powerpoint • 6.1-6.3 Quiz Fri
6.1-6.3 Review • AP Answers (even): • 2) D 14) E • 4) C 18) A • 6) C 20) B • 8) D 2a) • 10) C b) • 12) C c)
6.1-6.3 Review • 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 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!