# Area of solid revolution using integration. [duplicate]

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 ?

• Consider reformatting your question with better grammar. Also, check out this link to see how to properly format the math portion of the question. This will improve readability and increase the likelihood of receiving help. meta.math.stackexchange.com/questions/5020/… Commented Jul 25, 2014 at 22:59
• I assume you're talking about surface area. The short answer is that the angle of the surface matters. It's similar to the reason that we need to take into account slope when we calculate arc length by integration. Commented Jul 25, 2014 at 23:13
• why we didn't do the same to volume , and considered the angle of the surface ? is n't the same concept? :) Commented Jul 25, 2014 at 23:15
• No, although unfortunately I can't really formalize why (maybe someone with a stronger background can help). The best I can say is that in volume calculations the angle becomes insignificant approaching infinite disks, but this isn't the case for surface area. The best analogy I can come up with (this is sort of the same issue but "down a dimension," if that makes sense) is the classic false proof that $\pi=4$ (math.stackexchange.com/questions/12906/is-value-of-pi-4). Indeed, the area enclosed by the jagged outline approaches the area of the circle, but the perimeter doesn't go to $\pi$. Commented Jul 25, 2014 at 23:23

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.