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8867564
Group Title
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
 2 years ago
8867564 Group Title
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
 2 years ago

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

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

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

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

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

colorful Group TitleBest ResponseYou've already chosen the best response.0
dw:1337146393852:dw
 2 years ago

colorful Group TitleBest ResponseYou've already chosen the best response.0
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\[A(x)=2\pi rh=2\pi x^2\cdot x=2\pi x^3\]
 2 years ago

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

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

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

timo86m Group TitleBest ResponseYou've already chosen the best response.0
deos y=x^2 or x*2?
 2 years ago

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

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

timo86m Group TitleBest ResponseYou've already chosen the best response.0
I got 127.5000000*Pi
 2 years ago

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

marco26 Group TitleBest ResponseYou've already chosen the best response.0
what is the answer?
 2 years ago

timo86m Group TitleBest ResponseYou've already chosen the best response.0
I got 127.50*Pi
 2 years ago
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