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Finding the volume of a cylinder

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    x2+y2+16

    2. Relevant equations

    [tex]\pi[/tex][tex]\int^{b}_{a}[/tex]{R(x)2}dx

    3. The attempt at a solution

    I just need to know if i set this up right.

    When I solve for y and graph it I get a semi circle that goes from -4 to 4.

    [tex]\pi\int^{0}_{-4}[/tex]{([tex]\sqrt{(16-x^{2})}[/tex])[tex]^{2}[/tex]}dx + [tex]\pi\int^{4}_{0}[/tex]{([tex]\sqrt{(16-x^{2})}[/tex])[tex]^{2}[/tex]}dx

    I get 85.4[tex]\pi[/tex]
     
  2. jcsd
  3. Feb 1, 2009 #2
    You have given an expression but i'll assume you mean the equation for:

    [itex] x^2 + y^2 - 16 = 0[/itex]

    which is a circle whose centre is at 0,0 and has a radius of 4 units?

    The area of such a cylinder would be

    [itex] 16\pi l [/itex]

    where l is the length of the cylinder.

    I don't think you need to use [itex] \int_a^b y^2 dx [/itex] because you know the radius and it's not a volume of revolution.

    If you want to find the sphere when the shape is rotated about the y axis pi radians you can just use the formula

    [itex] \frac{4}{3} \pi r^3 [/itex]

    [itex] \frac{256}{3} \pi [/itex]

    [itex] 85.3 \pi [/itex]

    So no need for integration, but you got the right answer.
     
  4. Feb 1, 2009 #3

    Dick

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    Science Advisor
    Homework Helper

    It looks to me like you are trying to find the volume of a SPHERE by rotating a semicircle. So yes, the answer is correct. But no need to round it off or to break the integral into two parts. Like Gregg said it's pi*256/3. And the equation is x^2+y^2=16. This has nothing to do with cylinders.
     
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