Washer Method and Shell Method setup issues

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Homework Help Overview

The problem involves finding the volume of a solid formed by revolving a specific area defined by the equation x^(2/3) + y^(2/3) = a^(2/3) around the line x = a. The original poster seeks assistance with setting up the problem for both the washer method and the cylindrical shell method.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster expresses confusion about how to set up the problem, particularly regarding the use of the constant 'a' in the context of the washer method and cylindrical shell method. Participants question the necessity of solving for 'a' and discuss the implications of using 'dx' versus 'dy' for the volume slices.

Discussion Status

Participants are actively engaging with the original poster's questions, providing clarifications about the methods and encouraging further exploration of the setup. There is a lack of consensus on the specific limits and the interpretation of 'a' in the context of the problem.

Contextual Notes

Participants note that the original poster is new to the forum and may be unfamiliar with the conventions of using constants in volume calculations. There is an emphasis on understanding the geometric interpretation of the methods being discussed.

jonwill
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A solid is formed by revolving the area inside x^(2/3)+y^(2/3)=a^(2/3), a>0; to the right of x=0 and above y=0 about the line x=a. Find the Volume of the solid using;
A) The washer Method
B) Cylindrical Shell Method

I am at a loss at how to set up this problem...I understand how to do the two methods once this problem is set up but I am just unable to do so.

I was able to solve for a and got a=(x^(2/3)+y^(2/3))^(3/2), but from there I am lost. I am new to the forum and only wished I had found it sooner. Everyone here seems very bright and knowledgeable.

Thanks for your time,
Jon
 
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Welcome to PF!

Hi Jon! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

Why are you solving for a? a is a constant :confused:

Both the washer method and the cylindrical shell method involve slicing the volume into easy-to-calculate regions.

The washer method uses "horizontal" slices, each has height dy and is shaped like a washer, ie it has an inner radius and an outer radius, so find the radii of each slice as a function of y, then find the volume (it will be a multiple of dy, the height), and integrate.

The cylindrical shell method uses "cookie-cutter" slices, each has thickness dx and is shaped like the round skin of a cylinder, so find the height of each slice as a function of x, then find the volume (it will be a multiple of dx, the thickness), and integrate.

What do you get? :smile:
 
Ok Thank you,
Solving for y:
y=(a2/3-x2/3)3/2

How would I plug this into the formula for the washer method? I get very confused when I am only given an "a" constant to work with instead of a solid number.

V=[tex]\Pi[/tex][tex]\int[/tex]ba((R(x))2-(r(x))2)dx
 
Hi Jon! :smile:

(have a pi: π and an integral: ∫ :wink:)
jonwill said:
Ok Thank you,
Solving for y:
y=(a2/3-x2/3)3/2

How would I plug this into the formula for the washer method? I get very confused when I am only given an "a" constant to work with instead of a solid number.

V=[tex]\Pi[/tex][tex]\int[/tex]ba((R(x))2-(r(x))2)dx

Isn't the outer radius always "a"? :confused:

And what are your limits "a" and "b" supposed to be?

And i see you're using dx …

what slice is dx the width of?​
 

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