Surface Area of y=(x)^1/2 Rotating About the x Axis

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Discussion Overview

The discussion revolves around finding the surface area of the curve defined by y = (x)^(1/2) when rotated about the x-axis, specifically for the interval 0 ≤ x ≤ 2. The focus is primarily on the mathematical reasoning and application of the surface area formula in this context.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant expresses difficulty in understanding how to calculate the surface area for the given function when rotated about the x-axis.
  • Another participant provides a formula for surface area, indicating it involves integrating a function that combines the curve's radius and arc length.
  • A participant seeks clarification on the meaning of "dx" in the context of the integral, indicating confusion about its role in the calculation.
  • There is a suggestion that "dy/dx" must be derived by integrating y with respect to x, though the specifics of this process are not fully explored.

Areas of Agreement / Disagreement

The discussion reflects a lack of consensus, as participants express differing levels of understanding regarding the integral and its components, particularly the interpretation of "dx" and "dy/dx."

Contextual Notes

Participants do not clarify the assumptions or definitions related to the integral, nor do they resolve the specific steps needed to evaluate the integral for the surface area.

mannaatsb
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i've been stuck on this problem for an hour now. how do you find the surface area for y=(x)^1/2 when 0_<x_<2, rotating about the x axis?
 
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You have a nice little formula for this kind of thing.

[tex]SA = \int_{x_i}^{x_f} 2\pi y(x) \sqrt{1 + \Big(\frac{dy}{dx}\Big)^2}\,dx[/tex]

You'll recognize that this is [itex]2\pi y(x)[/itex] times the formula for the arc length. Basically what you're doing is adding up a bunch of little rings of radius y and length an infintesimal piece of the arc length. Because the radius is y, and we know that circumference = [itex]2\pi r[/itex], that's where the [itex]2\pi y(x)[/itex] comes from. The "width" of the ring depends on y and x and can be made into a triangle of base 1 and height dy/dx.

cookiemonster
 
Last edited:
but how do you figure out what the dx part is? that's the part that i don't understand.
 
dx? dx is dx. It's necessary to evaluate the integral.

Do you mean dy/dx? You have to integrate y with respect to x to get dy/dx.

cookiemonster
 

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