Calculating surface area of a cone using integral. Probably this question would have been asked so many times. Why does the surface area value comes out to be incorrect when we use simple height instead of slant height when doing integral?
And why does this approach work when we calculate the volume?
Hope you understand what I am asking. Please help!!. :)
For further illustration you can click on this link.
 A: You don't need calculus to calculate the surface area of a cone.
Haven't you ever made one.  Imagine a paper cone without a base as shown below then imagine cutting it open as shown below.


Clearly the piece you will have must be an arc of radius $s$ and length $2 \pi r$ The circumference of the base.
So the surface area can be shown to be $\pi r s$ if we ignore the base. 
If you want to know how to calculate the volume take a look at my answer to another question here

Since you clearly stated that you want to use calculus to solve this problem imagine cutting the cone in slices parallel to the base then cut one of these slices.  What will its area be?
the length will be slightly shorter one side than the other but it will be $2 \pi r_x$ very nearly where $r_x$ is the radius of the slice you cut at x. Its width however will not be $\delta h$  the height of the slice you cut but $\delta s$ the slope height of the slice you cut.
When calculating the volume you want to add together the volumes of each slice.  The fact that the slice is chamfered does not change its area much each slice has volume $\pi r^2 h$. There is a good you tube video to illustrate this here
A: Using the vertical height to compute surface area can't possibly be correct. It would tell you that a cone with a height of almost zero would also have a surface area of almost zero. This is obviously not true for a big wide flat cone.
But things work differently when you compute volume -- a cone with a height of almost zero really does have a volume that's almost zero. This doesn't really prove that it's correct to use vertical height to compute volume. But, at least it doesn't have the same obvious flaw as the area calculation.
A: $$
V
=
\int_{V}{\rm d}V = {1 \over 3}\int_{V}\nabla\cdot\vec{r}\,{\rm d}V
=
{1 \over 3}\int_{S}\vec{r}\cdot{\rm d}\vec{S}
$$
Taking the vertex as the origin, you just need an integration over the base surface.
