# The Relationship between the Surface Area of a Curve and A Circle's Circumference

• Feb 9th 2011, 08:47 AM
JewelsofHearts
The Relationship between the Surface Area of a Curve and A Circle's Circumference
In my BC Calc class, we are learning about the formula for finding the surface area of curves of revolution. I don't like just memorizing formulas without understanding why they work, so I have set out to discover how the formula for the surface area of a curve was derived.

I'm concerned as to why the formula does not represent the sum of the circumference of the infinite number of circles formed by the curve, rather than the sum of surface area of an inifinite number of cylinders. When I solve problems using my hypothetical method I get a different answer than I would using the formula I have been taught.

• Feb 9th 2011, 09:20 AM
VonNemo19
Quote:

Originally Posted by JewelsofHearts
In my BC Calc class, we are learning about the formula for finding the surface area of curves of revolution. I don't like just memorizing formulas without understanding why they work, so I have set out to discover how the formula for the surface area of a curve was derived.

I'm concerned as to why the formula does not represent the sum of the circumference of the infinite number of circles formed by the curve, rather than the sum of surface area of an inifinite number of cylinders. When I solve problems using my hypothetical method I get a different answer than I would using the formula I have been taught.

We derive the formula for surface area in much the same way that we derived the formula for arc length. So, take a curve in the plane $\displaystyle y=f(x)$, and for simplicity assume that the curve exists in the first quadrant over the interval $\displaystyle [a,b]$. Begin by slicing the interval into subinterval of length $\displaystyle \Delta{x}$ and on each subinterval we will approximate the lentgh of the function by summing the straight lines. Then Rotate! That's it. The rotation genterates a bunch of surface areas of frustrums and the formula for the surface of a frustrum is $\displaystyle A_f=2\pi{r}l$ where $\displaystyle r$ is the average radius of the left hand radius and the right hand radius, or $\displaystyle r=\frac{1}{2}(r_1+r_2)$.

For the frustrum on the interval $\displaystyle [x_{i-1},x_i]$ we have $\displaystyle r_1=f(x_i)$ and $\displaystyle r_2=f(x_{i-1})$.

Now the length of the ith line segment is given by $\displaystyle \sqrt{1+[f'(x)]^2}\Delta{x}$. This can be derived and I can do it if you like. So, we have

$\displaystyle A_i=2\pi\left(\frac{f(x_i)+f(x_{i-1})}{2}\right)\sqrt{1+[f'(x)]^2}\Delta{x}$. But if delta x is small, we can say that $\displaystyle A_i=2\pi{f(c_i)}\sqrt{1+[f'(c_i)]^2}\Delta{x}$ for some $\displaystyle c_i\in[x_{i-1},x_i]$

Summing and taking the limit we get

$\displaystyle \displaystyle{A}=\lim_{n\to\infty}\sum_{i=1}^\inft y2\pi{f(c_i)}\sqrt{1+[f'(c_i)]^2}\Delta{x}=\int_a^b2\pi{f(x)}\sqrt{1+[f'(x)]^2}dx$
• Feb 9th 2011, 09:22 AM
Ackbeet
Incidentally, JewelsofHearts, you might want to make your language more precise. All space curves have zero surface area. (I don't think you were taking about space curves - perhaps surfaces?)
• Feb 9th 2011, 10:26 AM
VonNemo19
Ah, what the heck. I'll go ahead and derive the other part for ya.

OK. You know that if I break a smooth curve up into a bunch of little straight lines, then from two consecutive points on the curve, $\displaystyle P_{i-1}$ and $\displaystyle P_i$ a line can be drawn. The distance between these two points will be given by the distance formula like so

$\displaystyle d_{P_{i-1}P_i}=\sqrt{[f(x_{i})-f(x_{i-1})]^2+(x_i-x_{i-1})^2}$.

Now, changing notation for convenience, we let $\displaystyle f(x_{i})-f(x_{i-1})=\Delta{y}$ and $\displaystyle x_i-x_{i-1}=\Delta{x}$ sucht that

$\displaystyle d_{P_{i-1}P_i}=\sqrt{\Delta{y}^2+\Delta{x}^2}$

$\displaystyle =\sqrt{\Delta{x}^2(1+\frac{\Delta{y}^2}{\Delta{x}^ 2})}=\Delta{x}\sqrt{1+(\frac{\Delta{y}}{\Delta{x}} )^2}$.

But, $\displaystyle \frac{\Delta{y}}{\Delta{x}}=\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}=f'(c_i)$ for $\displaystyle c_i\in[x_{i-1},x_1]$ by the mean value theorem.

So therefore $\displaystyle d=\Delta{x}\sqrt{1+[f'(c_i)]^2}$
• Feb 10th 2011, 08:14 AM
JewelsofHearts
Wow, VonNemo19! That was a lengthy explanation. Thanks for taking the time to do that.:)

What I was really wondering, though, was if the surface area of a curve of revolution could be represented by the sum of the circumferences of an inifinite number of circles on a given interval.