Surface area of a curve

  • #1
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Homework Statement


An industrial settling pond has a parabolic cross section described by the equation [itex] y = \frac{x^2}{80} [/itex]. 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


Homework Equations


Length of a curve formula
[tex] ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2 } [/tex]


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:

[tex] s = \int_{-20}^{20} \sqrt{1 + \frac{x^2}{1600} } dx [/tex]
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...

Your input is greatly appreciated!
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


An industrial settling pond has a parabolic cross section described by the equation [itex] y = \frac{x^2}{80} [/itex]. 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


Homework Equations


Length of a curve formula
[tex] ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2 } [/tex]


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:

[tex] s = \int_{-20}^{20} \sqrt{1 + \frac{x^2}{1600} } dx [/tex]
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...

Your input is greatly appreciated!
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
STEMucator
Homework Helper
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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:
  • #4
LCKurtz
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Homework Statement


An industrial settling pond has a parabolic cross section described by the equation [itex] y = \frac{x^2}{80} [/itex]. 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


Homework Equations


Length of a curve formula
[tex] ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2 } [/tex]


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:

[tex] s = \int_{-20}^{20} \sqrt{1 + \frac{x^2}{1600} } dx [/tex]
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...

Your input is greatly appreciated!
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:
  • #5
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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
 

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