8867564
1) Find the volume of the solid of revolution formed when the region bounded between
x = 1, x = 4, y = x2 and y = 0
is revolved vertically around the x-axis.
Using cans is the answer 21π?
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timo86m
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by x2 you mean x*2 or x^2
timo86m
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if x*2
timo86m
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|dw:1337145858842:dw|
I think
timo86m
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area of a small slice will result in a ring so it be
pi r2^2 pi r1^2
colorful
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I know it as shell method
never heard "cans" befor lol
colorful
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|dw:1337146309981:dw|we need the height and radius of each cylinder...
colorful
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|dw:1337146393852:dw|
colorful
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the surface area of a cylinder is given by\[2\pi rh\]so in this case each surface area as a function of x is\[A(x)=2\pi rh=2\pi x^2\cdot x=2\pi x^3\]
colorful
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adding up all these areas from x=1 to x=4 should give us the correct integral
colorful
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\[V=2\pi\int_{1}^{4}x^3dx\]
8867564
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Thank you for the help on this problem. It was very helpful.
timo86m
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deos y=x^2 or x*2?
8867564
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It's x^2
timo86m
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oh
timo86m
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you can take area from y=0 to y=1
use
pi*4^2-pi*1^2=15 pi so that is first part
timo86m
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now for 1 to 16
integral(pi*4^2-pi(sqrt(y))^2dy) from 1 to 16
timo86m
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I got 127.5000000*Pi
timo86m
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let me know what the answer was :)
marco26
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what is the answer?
timo86m
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I got 127.50*Pi