Asked by Sean
The region between the graphs of x=y^2 and x=5y is rotated around the line y=5. What is the volume?
Answers
Answered by
oobleck
The lune-shaped region has vertices at (0,0) and (25,5)
So, the axis of rotation just touches the region. Using washers of thickness dx,
v = ∫[0,25] π(R^2-r^2) dx
where R = 5-5y = 5 - x/5 and r=5-y^2 = 5 - √x
v = ∫[0,25] π((5 - x/5)^2-(5 - √x)^2) dx = 625π/6
Using shells of thickness dy, we have
v = ∫[0,5] 2πrh dy
where r = 5-y and h = 5y-y^2
v = ∫[0,5] 2π(5-y)(5y-y^2) dy = 625π/6
So, the axis of rotation just touches the region. Using washers of thickness dx,
v = ∫[0,25] π(R^2-r^2) dx
where R = 5-5y = 5 - x/5 and r=5-y^2 = 5 - √x
v = ∫[0,25] π((5 - x/5)^2-(5 - √x)^2) dx = 625π/6
Using shells of thickness dy, we have
v = ∫[0,5] 2πrh dy
where r = 5-y and h = 5y-y^2
v = ∫[0,5] 2π(5-y)(5y-y^2) dy = 625π/6
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