Surface area enclosed by Cylinders

[RKD89]

I want to know how Surface areas enclosed by Cylinders , Cones..etc can be calculated using calculus integration...

 

[VooDooX]

well a cylinder is just a square that 2 ends have been touched so wouldn't 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 there's a thickness in which case R=distance to inner rim instead of outer this includes the top and bottom surface areas also

 

[HallsofIvy]

I want to know how Surface areas enclosed by Cylinders , Cones..etc can be calculated using calculus integration...

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
$$\int\int (f(x,y)- g(x,y))dydc= \int_{r= 0}^R\int_{\theta= 0}^{2\pi} (f(r cos(\theta),r sin(\theta))- g(r cos(\theta),r sin(\theta))) r dr d\theta$$

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 $R(z- h)= r$, and below by z= 0, is given by<br /> $$\int_{r= 0}^R\int_{\theta= 0}^{2\pi} z rdrd\theta= \int_{r= 0}^R\int_{\theta= 0}^{2\pi} h+ \frac{r}{R} rdrd\theta$$[/itex]