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Surface area enclosed by Cylinders

  1. Dec 3, 2009 #1
    I want to know how Surface areas enclosed by Cylinders , Cones..etc can be calculated using calculus integration...........
     
  2. jcsd
  3. Dec 3, 2009 #2
    well a cylinder is just a square that 2 ends have been touched so wouldnt it be
    2 pi r 2 + 2 pi r h or 2(pi r 2) + (2 pi r)* hand the inside and outside have the same surface area unless theres a thickness in which case R=distance to inner rim instead of outer this includes the top and bottom surface areas also
     
  4. Dec 3, 2009 #3

    HallsofIvy

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    In both of those, it would probably be best to use cylindrical coordinates.

    If, for example, your region is bounded by the cylinder [math]x^2+ y^2= R^2[/itex],
    with top and bottom given by z= f(x,y) and z= g(x,y), respectively, then the volume is given by
    [tex]\int\int (f(x,y)- g(x,y))dydc= \int_{r= 0}^R\int_{\theta= 0}^{2\pi} (f(r cos(\theta),r sin(\theta))- g(r cos(\theta),r sin(\theta))) r dr d\theta[/tex]

    The volume of the region bounded above by the cone [itex]R^2(z-h)^2= x^2+ y^2[/tex] which, in cylindrical coordinates is [itex]R(z- h)= r[/itex], and below by z= 0, is given by
    [tex]\int_{r= 0}^R\int_{\theta= 0}^{2\pi} z rdrd\theta= \int_{r= 0}^R\int_{\theta= 0}^{2\pi} h+ \frac{r}{R} rdrd\theta[/tex]
     
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