Solve Volume of Solid w/ Shell Method: 4x - x2 & x2

In summary, the conversation discusses using the shell method to find the volume of a solid given by two intersecting functions rotated around a specific line. The individual mentions finding the height and radius, and ultimately getting a different answer than the one provided in the book. The expert suggests correcting the height and finding the correct radius, which is 4-x.
  • #1
icesalmon
270
13

Homework Statement


use the shell method to find the volume of the solid given by y = 4x - x2 y = x2, about the line x = 4

The Attempt at a Solution


I have to find a height and a radius multiply them together and integrate them over some bounded region, fair enough. My functions intersect at x = 2 and x = 0, my bounds have been set. I have to find my height, which I believe to be "x", and my radius. I believe that to be my top function, 4x - x2 - my bottom function, x2 so I multiply them within my integrand to get 4x2 - 2x3 after integrating I get 4x3/3 - x4 / 2. Evaluating this expression at x = 2 I get 32/3 - 24/3 multiplied by 2pi to get 16pi/3. My book tells me it is 16pi so I went back and looked at this a few times and did the calculation over and I'm getting the same answer and I'm not sure where I'm getting this wrong with my geometry.
 
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  • #2
icesalmon said:

Homework Statement


use the shell method to find the volume of the solid given by y = 4x - x2 y = x2, about the line x = 4

The Attempt at a Solution


I have to find a height and a radius multiply them together and integrate them over some bounded region, fair enough. My functions intersect at x = 2 and x = 0, my bounds have been set. I have to find my height, which I believe to be "x", and my radius. I believe that to be my top function, 4x - x2 - my bottom function, x2 so I multiply them within my integrand to get 4x2 - 2x3 after integrating I get 4x3/3 - x4 / 2. Evaluating this expression at x = 2 I get 32/3 - 24/3 multiplied by 2pi to get 16pi/3. My book tells me it is 16pi so I went back and looked at this a few times and did the calculation over and I'm getting the same answer and I'm not sure where I'm getting this wrong with my geometry.

Your height is 4x-2x^2. Your radius isn't x. You are rotating around x=4. What is it?
 
  • #3
*sigh* 4-x
 
  • #4
icesalmon said:
*sigh* 4-x

Riiighht.
 

What is the shell method?

The shell method is a mathematical technique used to find the volume of a solid of revolution. It involves slicing the solid into thin cylindrical shells and integrating their volumes to find the total volume of the solid.

What is the formula for the shell method?

The formula for the shell method is V = ∫2πrh dx, where r is the distance from the axis of revolution to the shell, h is the height of the shell, and dx is the thickness of the shell.

How do you solve for the volume of a solid using the shell method?

To solve for the volume of a solid using the shell method, you first need to identify the axis of revolution and the limits of integration. Then, you need to set up the integral using the formula V = ∫2πrh dx, where r and h are expressed in terms of the variable of integration. Finally, you integrate the function and evaluate the integral to find the volume.

What is the difference between the shell method and the disk method?

The shell method and the disk method are both techniques used to find the volume of a solid of revolution. The main difference between them is the shape of the slices used to approximate the solid. The shell method uses cylindrical shells, while the disk method uses circular disks. The choice between the two methods depends on the shape of the solid and the ease of integration.

Can the shell method be used to find the volume of any solid of revolution?

No, the shell method can only be used to find the volume of solids of revolution with a vertical axis of revolution. If the axis of revolution is horizontal, the disk method must be used instead.

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