Hello. I just started to study integral calculus not long ago and I have some confusion when it comes to calculating the areas of surfaces of revolution using integral.(adsbygoogle = window.adsbygoogle || []).push({});

As from my testbook, when we want to calculate this kind of surface area, we often use the frustum surface area to approximate it in ds slant. Then integrate it along the defining curve( f(x) ) as the following formular states:

S=∫2πf(x)ds=∫2πf(x)√(1+f'(x)^{2}) dx

However my problem is why we don't use the cylinder slice with dx thick to approximate it and then integrate over the specified region, as we do when calculating the volumes of solids of revolution.

And I found in this case the formular will just be:

S'=∫2πf(x) dx which is different from above.

So please someone tell me where I made a mistake. I am sorry my post is lengthy and I also wish I could explain clearer.

Thanks a lot for your time.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Areas of surfaces of revolution

Loading...

**Physics Forums | Science Articles, Homework Help, Discussion**