Surface Area of a Solid of Revolution

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Homework Help Overview

The discussion revolves around finding the surface area generated by rotating the parabola defined by \(y = x^2\) around the y-axis, specifically for the interval where \(0 \leq x \leq \sqrt{k}\). Participants are tasked with deriving a formula in terms of the constant \(k\).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the surface area formula \(S = 2\pi \int y \, ds\) and expresses a belief that the resulting formula may relate to the volume of a sphere. Other participants question the setup of the problem, particularly the choice of variable for the radius of rotation and the implications of rotating around the y-axis.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the correct setup for the rotation. There is an exploration of the relationship between surface area and volume, with no clear consensus yet on the correct approach or interpretation of the problem.

Contextual Notes

There are indications of confusion regarding the variables used in the setup, particularly the distinction between using \(x\) and \(y\) in relation to the axis of rotation. This may affect the interpretation of the surface area calculation.

Ki-nana18
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Homework Statement


Find the area of the surface generated when you rotate the parabola y=x2 0 less than or equal to x less than or equal to the square root of k, around the y-axis. You should end up with a simple formula in terms of the constant k.


Homework Equations


S=2[tex]\pi[/tex][tex]\int[/tex]yds

The Attempt at a Solution


I suspect that the simple formula is the volume of a sphere. I got all the way to applying the fundamental theorem of calculus and so far I have 2pi[((12(square root of k)+3)/(18))^(3/2)-(1/12)]
 
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Ki-nana18 said:

Homework Statement


Find the area of the surface generated when you rotate the parabola y=x2 0 less than or equal to x less than or equal to the square root of k, around the y-axis. You should end up with a simple formula in terms of the constant k.


Homework Equations


S=2[tex]\pi[/tex][tex]\int[/tex]yds

The Attempt at a Solution


I suspect that the simple formula is the volume of a sphere. I got all the way to applying the fundamental theorem of calculus and so far I have 2pi[((12(square root of k)+3)/(18))^(3/2)-(1/12)]

Your setup is incorrect because the radius of rotation should be x, not y. But I'm curious why you would think the answer for surface area would give a volume? And of a sphere?
 
Well I figured since I am rotating around the y-axis I would have the equation set up as x=[tex]\sqrt{}y[/tex].
 
LCKurtz said:
Your setup is incorrect because the radius of rotation should be x, not y. But I'm curious why you would think the answer for surface area would give a volume? And of a sphere?

Ki-nana18 said:
Well I figured since I am rotating around the y-axis I would have the equation set up as x=[tex]\sqrt{}y[/tex].

When you rotate about the y-axis the radius is in the x direction, your equation is given in terms of x, and the natural variable to use is x.
 

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