Asked by Dib
use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. xy=1, x=0, y=1, y=3
Answers
Answered by
Steve
Using shells, you have thickness dy, so
v = ∫[1,3] 2πrh dy
where r = y and h = x = 1/y
v = 2π∫[1,3] y*1/y dy
= 2π∫[1,3] dy
= 2πy [1,3]
= 2π(3-1)
= 4π
That strikes me as odd. Let's try discs (washers) of thickness dx. The volume on [0,1/3] is just a hollow cylinder, of volume (1/3)π(3^2-1^2) = 8/3 π.
The remaining part has volume
v = ∫[1/3,1] π(R^2-r^2) dx
where R=y=1/x, r=1
v = π∫[1/3,1] (1/x^2-1) dx
= π(-1/x-x)[1/3,1]
= π((-1-1)-(-3-1/3))
= π(-2+10/3)
= 4/3 π
Add to that the 8/3 π from the constant part, and you have 4π
v = ∫[1,3] 2πrh dy
where r = y and h = x = 1/y
v = 2π∫[1,3] y*1/y dy
= 2π∫[1,3] dy
= 2πy [1,3]
= 2π(3-1)
= 4π
That strikes me as odd. Let's try discs (washers) of thickness dx. The volume on [0,1/3] is just a hollow cylinder, of volume (1/3)π(3^2-1^2) = 8/3 π.
The remaining part has volume
v = ∫[1/3,1] π(R^2-r^2) dx
where R=y=1/x, r=1
v = π∫[1/3,1] (1/x^2-1) dx
= π(-1/x-x)[1/3,1]
= π((-1-1)-(-3-1/3))
= π(-2+10/3)
= 4/3 π
Add to that the 8/3 π from the constant part, and you have 4π
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