# Question about Integrals to Determine Volume vs Surface Area

• I
• Ascendant0
Ascendant0
I'm a little thrown off with material I'm going through right now. I already covered the whole "area under the curve" and using that to determine the volume of a given equation, but I'm confused now as to why calculating the surface area has a different method with ds?

For example, say there is a trumped-shaped object whose outer shell is defined by ## y=x^2 ## that spins around the x-axis from x = 0 to x = 3. If I want to find the volume within that object, it would be:

## \int_{0}^{3} \pi x^4 dx ## // I believe this is correct, basically the integral of the area of all "pieces" of circles created by dx from 0 to 3, right?

Since dx gives the location of the outside of that object at any given point, if I wanted to then calculate the surface area of that same object, my initial thoughts were you simply do the same thing, but for the circumference of the circle (## 2 \pi r ##) applied to the equation, giving:

## \int_{0}^{3} 2 \pi x^2 dx ## // I feel since dx gives the location of the outer edge at any point, simply calculate the total of the outer edge's circumference from 0 to 3, and that would be it

Now, since I've already made this lengthy, in short, I have seen where instead of "dx" it is "ds" and a different way to calculate it, but I'm not sure why? Can someone please help me to visualize this? I feel like dx would give you the infinitesimal point at any given spot between 0 to 3, so if you just calculate the circumference for everything from 0 to 3, that would give you the correct answer. Clearly, there' s a part of this that I'm not visualizing correctly, but I can't figure out what?

You need the arclength along $(x, f(x))$ multiplied by the circumference of the circle. In terms of small increments the arclength is given by Pythagoras's theorem, $$\delta s^2 = \delta x^2 + \delta y^2 \approx (1 + f'(x)^2)\delta x^2$$ which in the limit and taking square roots becomes $$ds = \sqrt{1 + f'(x)^2}\,dx.$$

bhobba and Ascendant0
pasmith said:
You need the arclength along $(x, f(x))$ multiplied by the circumference of the circle. In terms of small increments the arclength is given by Pythagoras's theorem, $$\delta s^2 = \delta x^2 + \delta y^2 \approx (1 + f'(x)^2)\delta x^2$$ which in the limit and taking square roots becomes $$ds = \sqrt{1 + f'(x)^2}\,dx.$$
I know that's why it's done, but what I don't get is why there is the need, since with the integral, it essentially breaks it down into infinitesimal parts anyway. With an infinitesimal piece, I'm thinking all that should really matter is the location of that bit of it. That's what I'm having a hard time understanding. I would get it if we were measuring something on a larger scale and there was an arclength, but when broken down as an integral (and the limits concept behind them), doesn't it become infinitesimal, so that dimensions (like curvature of an arc) become irrelevant?

Maybe the picture on the first page of this PDF will help you visualize it:
https://www.lboro.ac.uk/media/media...HELM Workbook 29 Integral Vector Calculus.pdf

Or for those who can't click on it, it is the following:

bhobba, Lnewqban and Ascendant0
Thank you for the help. Yes, it makes sense to me now. It should've days ago, lol. Just been extremely sick lately, so I was having a hard time processing it. I appreciate you, thanks!

Lnewqban
Ascendant0 said:
Thank you for the help. Yes, it makes sense to me now. It should've days ago, lol. Just been extremely sick lately, so I was having a hard time processing it. I appreciate you, thanks!

If you rotate the line ##y=x## instead I think it becomes easier to visualize. As you integrate from 0 to 3 you act as if each little slice of ##dx## is a horizontal wall, which underestimates how long it really is.

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