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 \(\displaystyle 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
$$\int_{r= 0}^R\int_{\theta= 0}^{2\pi} z rdrd\theta= \int_{r= 0}^R\int_{\theta= 0}^{2\pi} h+ \frac{r}{R} rdrd\theta$$\)