Shell Method Problem: Solve x=y^(2), x=4 About the X-Axis

In summary, the conversation discusses using the shell method to solve a problem involving the equation x=y^(2), x=4, about the x axis. The attempt at a solution involves using the integral from -2 to 2 of y*(4-y^(2)) dy, but the answer comes out to be 0 instead of the expected 8pi. The conversation also touches on another related problem and explains the reasoning behind the chosen limits for the integral.
  • #1
hvidales
29
0

Homework Statement


x=y^(2), x=4, about the x axis.

Homework Equations



2pi* integral from a to b of radius*height of function*thickness

The Attempt at a Solution


I have 2pi* integral from -2 to 2 of y*(4-y^(2)) dy but that does not make any sense. Answer comes out to be 0. The real answer is 8pi. I know how to do this with the disk method just not the shell method. Thank for the help!
 
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  • #2
hi hvidales! :smile:
hvidales said:
… integral from -2 to 2 …

noooo :wink:
 
  • #3
I see and thanks. However, how come for this problem: y=x^(2), y=2-x^(2), about x=1 the limits are from -1 to 1 ?
 
  • #4
you mean x goes from -1 to 1?

that's because the body in that case goes from x = -1 to x = 3 (the reflection about x = 1)

in your first example the body went from y = -2 to y = 2

in each case, you're taking half (because each single cylindrical shell is in both halves) :wink:
 
  • #5
Thank you!
 

FAQ: Shell Method Problem: Solve x=y^(2), x=4 About the X-Axis

What is the Shell Method problem and how do I solve it?

The Shell Method problem is a type of integration problem in calculus that involves finding the volume of a solid of revolution. In this specific problem, we are given two equations, x=y^(2) and x=4, and we need to find the volume of the solid formed when these two curves are rotated around the x-axis. To solve this problem, we will use the Shell Method formula, which involves integrating the circumference of a cylindrical shell over a certain interval.

What is the difference between the Shell Method and the Disk Method?

The Shell Method and the Disk Method are both methods used to find the volume of a solid of revolution. The main difference between them is the shape of the cross-sections used to approximate the volume. In the Shell Method, we use cylindrical shells, while in the Disk Method, we use circular disks. The Shell Method is typically used when the axis of rotation is parallel to the axis of integration, while the Disk Method is used when the axis of rotation is perpendicular to the axis of integration.

How do I set up the integral for the Shell Method problem?

To set up the integral for the Shell Method problem, we first need to determine the limits of integration. This can be done by finding the points of intersection between the two given equations, x=y^(2) and x=4. Next, we need to determine the height of each cylindrical shell, which is given by the difference between the two equations. Finally, we use the Shell Method formula: V = 2π∫(radius)(height)dx, where the radius is the distance from the axis of rotation to the shell and dx represents the thickness of each shell.

Can I use the Shell Method for any solid of revolution?

No, the Shell Method can only be used for solids of revolution that have a cylindrical shape. This means that the axis of rotation must be parallel to the axis of integration. If the axis of rotation is perpendicular to the axis of integration, we must use the Disk Method instead.

Are there any other methods I can use to solve this type of problem?

Yes, in addition to the Shell Method and the Disk Method, there is also the Washer Method. The Washer Method is similar to the Disk Method, but instead of using circular disks, we use circular washers (or annuli) to approximate the volume. The Washer Method is typically used when the region between the two curves is not a solid shape, such as when the curves intersect multiple times or when there are holes in the solid of revolution.

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