# Surface area of a curve

1. Aug 8, 2013

### stunner5000pt

1. The problem statement, all variables and given/known data
An industrial settling pond has a parabolic cross section described by the equation $y = \frac{x^2}{80}$. the pond is 40 m across and 5 m deep at the cetner. the curved bottom surface of the pond is to be covered with a layer of clay to limit seepage from the pond. determine the surface are on the clay bottom of the pond

2. Relevant equations
Length of a curve formula
$$ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2 }$$

3. The attempt at a solution
I was thinking that if we simply find the length of the curve, then that would yield the surface area of the curve. This gives the following:

$$s = \int_{-20}^{20} \sqrt{1 + \frac{x^2}{1600} } dx$$
Which gives an answer of 66.9

The answer however is 1332.2 m^2. This leads me suspect that we must include the fact that this curve is to be rotated around the y axis making a bowl. I am not sure how to rotate this curve though...

2. Aug 8, 2013

### Ray Vickson

What is the formula for the surface area of the surface z = f(x,y)? (Here, my z = your y, and (x,y) are the planar coordinates on the pond's surface.)

3. Aug 8, 2013

### Zondrina

I read the question a few times over and I think I've got it. Plotting $y = \frac{x^2}{80} - 5$ on wolfram gives an accurate model of the problem ( The cross section of the pond which is 5m deep at the center ).

I believe what you would want to use here is :

$2 \pi \int_{c}^{d} x \sqrt{1 + (\frac{dx}{dy})^2} dy$ which will be the surface area of your curve when you rotate it about the y axis. The reason you want to rotate it about the y-axis is because you want the surface area of the ENTIRE cross section of the pond.

I got an answer of 1332.18.

Hint : Re-arrange your equation for y. Your limits should be from the bottom of the cross section of the pond to the top.

Last edited: Aug 8, 2013
4. Aug 8, 2013

### LCKurtz

You almost have it. The $ds$ element is at distance $x$ from the y axis, so the circumference of the circle it traces is $2\pi x$. You take that times the $ds$ length to get the surface area swept out. So you want $$\int_0^{20} 2\pi x~ds = \int_0^{20} 2\pi x\sqrt{1 + \frac{x^2}{1600} }~dx$$You only go from $0$ to $20$ in the limits because revolving it gets the "other side".

Last edited: Aug 8, 2013
5. Aug 8, 2013

### stunner5000pt

Thank you very much for all your help. I see now the integrand is the curved surface area of a cylinder where the radius is x and the height is the length of a curve formula