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Volume of solid in region

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data
    I want to convert this into polar and use double integral to find the volume of the solid in this region. I just need help setting this up
    region
    Q: x^2+y^2≤9, 0≤z≤4
    I know this is a cylinder with a height of 4.
    I am just having trouble incorporating this height into the integral.
    3. The attempt at a solution
    ∫_0^2π▒〖∫_0^3▒4 r〗 drd(theta)
    this is currently what I have
     
  2. jcsd
  3. Nov 4, 2012 #2
    "integral from 0 to 2pi" then integral 0 to 3. then 4 rdrdθ
     
  4. Nov 4, 2012 #3
    This is correct.
    There is a cylinder with height 4. When using a double integral to find the volume of a solid object, you can set it up with the "Top - Bottom" as the function to integrate. This can also be done by adding in a third integral and integrating 1.
    [itex] \int_0^{2\pi} \int_0^3 \int_0^4 (1)dV [/itex], where [itex]dV[/itex] is [itex]rdzdrd\theta[/itex].
    [itex] =\int_0^{2\pi} \int_0^3 (4) (r)drd\theta [/itex]
    You can also check this by using the formula for the volume of a cylinder which is [itex]\pi r^2h[/itex]
     
  5. Nov 4, 2012 #4
    thank you
     
  6. Nov 4, 2012 #5
    [itex]\int^\3_\0[/itex]
     
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