3

To find volume of the shape generated by rotating $f(x)$ around the $x$ axis we calculate

$$\int_a^b \pi f(x)^2\, \mathrm d x$$

as the area of a circle is $\pi r^2$ and we just split it into discs.

Why then by analogy is the surface area formula not just

$$\int_a^b 2 \pi f(x)\, \mathrm d x$$

I don't understand the fundamental difference between the two cases.

user85798
  • 1
  • 2
    I think the problem is the same as here –  Feb 25 '16 at 10:17
  • 1
    Yes I agree. As you approach the limit by making the discs thinner, you do indeed get a diagonal staircase whose surface area does not approach the actual surface area, although the volume of the shape does approach the actual volume. – Shuri2060 Feb 25 '16 at 10:22
  • that formula is only valid for a cylinder, it is not correct for anything more complicated, even a cone. You have to improve the approximation of each elemental disc in the integration by calculating the length of the circular edges, which involves the line integral. Try it out with the cone. – user247608 Feb 25 '16 at 11:09
  • So how do you justify that the formula is actually correct? How do I know at which point (in general) the approximation is good enough? – user85798 Feb 25 '16 at 11:11
  • @Sekots: Have you looked at the two duplicates pointed out? – joriki Feb 25 '16 at 11:38

0 Answers0