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Finding the Volume of a Solid of Revolution
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Consider two functions π and π bounding a region π·
For this example, π π₯ = 1 8 π₯ 4 and π π₯ = π₯ 3 β3 π₯ 2 +3π₯ Now imagine we rotate π· around the π₯ axis
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(Using Mathematica) We investigate the problem of finding the volume of this object
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Letβs consider a small slice of width ππ₯ of the region π·
We can find the volume when it is rotated around the x axis. Width=ππ₯ Area=π π π₯ 2 βπ π₯ 2 ππ=WidthΓArea = π π π₯ 2 βπ π₯ 2 ππ₯ Integrate the volume element to get volume. a b Volume= π π π π π₯ 2 βπ π₯ 2 ππ₯ Where [π,π] is the interval over which π· lies.
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Example: Let π· be the object bounded by the graphs of π¦= 1 8 π₯ 4 and π¦= π₯ 3 β3 π₯ 2 +3π₯. Let πΈ be the object obtained by rotating π· around the π₯-axis. Find the volume of πΈ. Solution We first sketch π·. We see that the interval is π,π =[0,2]. The formula for the volume of πΈ isβ¦ Volume= 0 2 π π₯ 3 β3 π₯ 2 +3π₯ 2 β π₯ ππ₯ = π β π₯ π₯ 7 7 β π₯ 6 +3 π₯ 5 β 9 2 π₯ 4 +3 π₯ = 88π 63 β
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Example 2: Consider the region π· in the first quadrant bounded by π¦= sin π₯ and π¦= cos π₯ as in the picture. Let πΈ be the solid obtained by rotating π· about the π¦-axis. Find the volume of πΈ. Consider a small sliver of π· of width ππ₯. Solution When rotated around the π¦-axis, it becomesβ¦ a cylindrical shell of radius π₯, height cos π₯ β sin π₯ and thickness ππ₯. Hence ππ=2πΓradiusΓheight ππ₯ =2ππ₯ cos π₯ β sin π₯ ππ₯ And the integration will go from 0 to π/4 since sin π₯= cos π₯ when π₯=π/4 Volume= 0 π/4 2ππ₯ cos π₯ β sin π₯ ππ₯ = 2π cos π₯ +π₯ cos π₯ β sin π₯ +π₯ sin π₯ 0 π/4 = π 2 π 2 β4
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Example 3 Let π· be the region between π¦=π₯+1 and π¦= π₯β1 2 . Find the volume of the solid obtained by revolving π· around the π₯-axis. Solution We draw a sketch. Now we find the volume of a small slice. ππ=AreaΓππ₯ =π π π₯ 2 βπ π₯ 2 ππ₯ π= 0 3 π π₯+1 2 β π₯β1 4 ππ₯ =π 3 π₯ 2 β 5 3 π₯ 3 + π₯ 4 β 1 5 π₯ = 72π 5
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Example Let π· be the region bounded by π¦= ln π₯ , π₯=1, π¦= ln 3 . Let πΈ be the solid obtained by rotating π· about the line π₯=3. Find the volume of πΈ. Solution We draw a sketch. We pick a sliver of width ππ₯. Now we observe what happens to the sliver as it is rotated about the line π₯=3. We obtain a cylindrical shell with volume: ππ=2πΓRadiusΓHeightΓThickness =2π 3βπ₯ ln 3 β ln π₯ ππ₯ π= 1 3 ππ = 1 3 2π 3βπ₯ ln 3 β ln π₯ ππ₯ = 1 3 2π π₯β3 ln π₯ 3 ππ₯ Integration by parts = 6ππ₯β π π₯ 2 2 β6ππ₯ ln π₯ 3 +π π₯ 2 ln π₯ =8πβ5π ln 3
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See sections 7.2 and 7.3 in Stewart for problems on finding the volumes of solids of revolution.
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