1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook