# Surface area formula

I'm kind of confused about the explanation of the surface area formula in my text book

The text gave us $$\int_{a}^{b}2\pi f(x) \sqrt{1+[f'(x)]^2}dx$$ after that the formula is getting like $$\int_{a}^{b}2\pi y \sqrt{1+\left(\frac{dy}{dx}\right)^2 }dx$$

also it said $$\int_{a}^{b}2\pi y \sqrt{1+ \left(\frac{dx}{dy}\right)^2 }dy$$

They used $y$ instead of $f(x)$. It confused me.

How can I figure out between when I plug in some $y$ function at $y$ and when I change any thing (just use $y$) in the formula?

Kind of difficult to explain though, I hope you can understand my question. And if you have any idea, please post it. Thank you!

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Somehow I imagine it would be easier to respond to this if we could look at the actual text. – anon Feb 27 '12 at 3:45
Hi!! anon I think I can take a scan of textbook and share to this page. but i'm out now so as soon as i got home i will add the link of the text book copy – Ryu Feb 27 '12 at 3:52
You can probably just use Google Books. It would make your life a bit easier, along with ours. – Joe Feb 27 '12 at 3:58

## 2 Answers

It is not that hard.

The first formula is given for $$y = f(x)$$

So, you just choose a function and replace both $f(x)$ and $f'(x)$.

The second formula is simply using the $y$ notation instead of $f(x)$. So there you have again $$y = f(x)$$ and they write $$f'(x) = \frac{dy}{dx}$$

You can interpret the last formula as follows: suppose the $y$ axis is the independent variable and $x$ is the dependent variable, then

$$\int\limits_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy}$$

gives the surface of revolution around the $y$ axis - here we have $x=f(y)$.

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The last one made me confuse a lot actually. so Now I found a typo of text book then hehe your formula makes more sense for me !! thanks a lot ! – Ryu Feb 27 '12 at 4:11
Hmm. Are you sure about that last one? That is the integral you can use for computing the area of a surface of revolution, when the revolving is done about the $y$-axis. The other integrals calculate the area, when the curve is revolving about the $x$-axis. The factor outside the square root should be proportional to the radius of that tiny "washer/drum/disk/whatever". – Jyrki Lahtonen Feb 27 '12 at 5:50
@JyrkiLahtonen I'll update that. I thought they all referred to one formula and its variations in notation. – Pedro Tamaroff Feb 27 '12 at 23:05
Wait a minute, why does your last integral formula of 2pi y sqrt(...) instead of 2pi x sqrt(...) – thang Apr 10 '15 at 6:38

What you are looking at is the formula for the area of the surface generated by revolving the curve y = f(x) about the x-axis. There are four parts to the formula a = lower bound of integration, b = upper bound of integration (the interval on the x-axis which the function is being defined), y = f(x) (the actual function being rotated that is in terms of x, and dy/dx (the derivative of the formula being rotated).

The formula for rotations about the y-axis will use the dx/dy and x=g(y) (a function defined in terms of y).

I think you are getting the two of them mixed up.

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