Area of solid revolution using integration. When we calculate the volume of a solid generated by rotating a curve around $x$-axis, We divide it into disks.
So ,we get  $dv = \pi  r^2 dx$. where $r=y$ and then we integrate. 
That OK, but when we calculate the area of the outer\external surface of the solid,  why don't we let $da=2 \pi r dx$ , and then integrate to get the area?  why is it wrong? note that   $r=y$. 
Any intuitive insights to tackle this ? 
 A: To illustrate intuitively why this doesn't and shouldn't work, think of function on the interval $[0,L]$ rotated around the x-axis. It should be clear that the area of the surface of revolution should depend on the length of the curve in that interval.
So $y = c$, for some constant $c$, has a length of L, and the surface generated is the side of a right circular cylinder. But for $y = cx$, an inclined line, the surface generated is a cone of height $L$.
It is also easy to see that the cone generated by $y = x$ has a smaller surface area than, say, the one generated by $y = 10x$, which is much wider. So the arc length of the curve that is revolved around the axis needs to be considered.
If you approximate the curve by straight horizontal line segments, no matter how many pieces you use the total length of all pieces will still be the length of the interval, $L$. So in that naive method, you are not really considering the key property of the shape that the area of the surface of revolution depends on.
