Question
Find the volume generated by revolving the region bounded by x=1, y= e^x/2, y= e^x about the x-axis?
Using the washer method or would it be the shell method. I'm having trouble finding that the boundaries would be? Thank you.
Using the washer method or would it be the shell method. I'm having trouble finding that the boundaries would be? Thank you.
Answers
I assume you mean e^(x/2)
You can use either method. Using washers of thickness dx, the ring of each washer is just the distance between the curves. Both graphs start at (0,1), so
v = ∫[0,1] π(R^2-r^2) dx
where R=e^x and r=e^(x/2)
v = ∫[0,1] π((e^x)^2-(e^(x/2))^2) dx
you can simplify that to
v = ∫[0,1] π(e^(2x) - e^x) dx = π/2 (e-1)^2
You can use either method. Using washers of thickness dx, the ring of each washer is just the distance between the curves. Both graphs start at (0,1), so
v = ∫[0,1] π(R^2-r^2) dx
where R=e^x and r=e^(x/2)
v = ∫[0,1] π((e^x)^2-(e^(x/2))^2) dx
you can simplify that to
v = ∫[0,1] π(e^(2x) - e^x) dx = π/2 (e-1)^2
What about the x=1 part does that just create the boundary or do I shift the graph to the y-axis[ left 1]?
the line x=1 sets the right limit of integration. The region is roughly triangular, with vertices at
(0,1), (1,√e), (1,e)
integrating in the x direction just stacks up all those washers horizontally.
You can use shells of thickness dy, but it gets a bit more complicated, since the right-hand boundary changes at (1,√e). From y=1 to y=√e, the height of the horizontal cylinders is the distance between the two curves
x = 2lny and x = lny
From y=√e to y=e, the height is just 1-lny
The volume of a shell is 2πrh, but h changes as we pass y=√e
v = ∫[1,√e] 2πy(2lny-lny) dy + ∫[√e,e] 2πy(1-lny) dy
you have to use integration by parts to get
∫y lny dy = y^2/4 (2lny-1)
Putting all that together, we have the final result of
v = 2π(y^2/4(2lny-1))[1,√e] + 2π(y^2/2 - y^2/4 (2lny-1))[√e,e]
Yes, I did the algebra, and it again comes out π/2 (e-1)^2
go with washers!
(0,1), (1,√e), (1,e)
integrating in the x direction just stacks up all those washers horizontally.
You can use shells of thickness dy, but it gets a bit more complicated, since the right-hand boundary changes at (1,√e). From y=1 to y=√e, the height of the horizontal cylinders is the distance between the two curves
x = 2lny and x = lny
From y=√e to y=e, the height is just 1-lny
The volume of a shell is 2πrh, but h changes as we pass y=√e
v = ∫[1,√e] 2πy(2lny-lny) dy + ∫[√e,e] 2πy(1-lny) dy
you have to use integration by parts to get
∫y lny dy = y^2/4 (2lny-1)
Putting all that together, we have the final result of
v = 2π(y^2/4(2lny-1))[1,√e] + 2π(y^2/2 - y^2/4 (2lny-1))[√e,e]
Yes, I did the algebra, and it again comes out π/2 (e-1)^2
go with washers!
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