anonymous
  • anonymous
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?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
by x2 you mean x*2 or x^2
anonymous
  • anonymous
if x*2
anonymous
  • anonymous
|dw:1337145858842:dw| I think

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anonymous
  • anonymous
area of a small slice will result in a ring so it be pi r2^2 pi r1^2
anonymous
  • anonymous
I know it as shell method never heard "cans" befor lol
anonymous
  • anonymous
|dw:1337146309981:dw|we need the height and radius of each cylinder...
anonymous
  • anonymous
|dw:1337146393852:dw|
anonymous
  • anonymous
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\[A(x)=2\pi rh=2\pi x^2\cdot x=2\pi x^3\]
anonymous
  • anonymous
adding up all these areas from x=1 to x=4 should give us the correct integral
anonymous
  • anonymous
\[V=2\pi\int_{1}^{4}x^3dx\]
anonymous
  • anonymous
Thank you for the help on this problem. It was very helpful.
anonymous
  • anonymous
deos y=x^2 or x*2?
anonymous
  • anonymous
It's x^2
anonymous
  • anonymous
oh
anonymous
  • anonymous
you can take area from y=0 to y=1 use pi*4^2-pi*1^2=15 pi so that is first part
anonymous
  • anonymous
now for 1 to 16 integral(pi*4^2-pi(sqrt(y))^2dy) from 1 to 16
anonymous
  • anonymous
I got 127.5000000*Pi
anonymous
  • anonymous
let me know what the answer was :)
marco26
  • marco26
what is the answer?
anonymous
  • anonymous
I got 127.50*Pi

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