Washer Method and Shell Method setup issues

In summary: The outer radius is always "a". a and b are the inner radius and outer radius of the washer-shaped slice, respectively.
  • #1
jonwill
2
0
Question)
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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  • #2
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:
 
  • #3
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
 
  • #4
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?​
 

1. What is the difference between the Washer Method and Shell Method?

The Washer Method and Shell Method are two different methods used to find the volume of a solid of revolution. The main difference between the two methods is in how they are set up. The Washer Method uses cylindrical shells to find the volume, while the Shell Method uses horizontal or vertical slices.

2. How do I determine the axis of revolution for a solid?

The axis of revolution is the line around which the solid is rotated to create a three-dimensional shape. This line can be determined by looking at the axis of symmetry of the two-dimensional shape being rotated or by considering the direction of rotation specified in the problem.

3. What are some common mistakes when setting up Washer Method and Shell Method problems?

Some common mistakes when setting up these types of problems include not properly identifying the axis of revolution, not correctly choosing the limits of integration, and not accounting for any holes or gaps in the solid being rotated.

4. How do I know which method to use for a given problem?

The method used to find the volume of a solid of revolution depends on the shape of the solid and the axis of rotation. If the axis of rotation is parallel to the axis of symmetry of the shape, then the Washer Method is used. If the axis of rotation is perpendicular to the axis of symmetry, then the Shell Method is used.

5. Can the Washer Method and Shell Method be used for any shape?

Yes, these methods can be used for any shape as long as it can be revolved around an axis to create a three-dimensional solid. However, for more complex shapes, it may be necessary to break the solid into smaller, simpler shapes in order to apply these methods.

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