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 2 years ago
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 xaxis.
Using cans is the answer 21π?
 2 years ago
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 xaxis. Using cans is the answer 21π?

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timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0by x2 you mean x*2 or x^2

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1337145858842:dw I think

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0area of a small slice will result in a ring so it be pi r2^2 pi r1^2

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0I know it as shell method never heard "cans" befor lol

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1337146309981:dwwe need the height and radius of each cylinder...

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0the 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
 2 years ago
Best ResponseYou've already chosen the best response.0adding up all these areas from x=1 to x=4 should give us the correct integral

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0\[V=2\pi\int_{1}^{4}x^3dx\]

8867564
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you for the help on this problem. It was very helpful.

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0you can take area from y=0 to y=1 use pi*4^2pi*1^2=15 pi so that is first part

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0now for 1 to 16 integral(pi*4^2pi(sqrt(y))^2dy) from 1 to 16

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0let me know what the answer was :)
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