Calculating Volume of Solid Using Cylindrical Shell Method

In summary, the conversation revolved around finding the volume of the solid obtained by rotating the region bounded by the curves y=x^4 and y=1 about the line y=7 using the cylindrical shell method. The formula for the method was mentioned and the integral was set up, but there were some initial errors in the boundaries and signs. The correct integral should be wtr y (from 0 to 1) with a radius of 7-y^{1/4}. Further assistance was offered to set up the integral correctly.
  • #1
pattiecake
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0
"simple" shell

I know this is relatively simple, but I'm a little rusty. Could someone help me out? We want to find the volume of the solid obtained by rotating the region bounded by the curves y=x^4 and y=1 about the line y=7 using the cylindrical shell method.

According to my book the general formula for cylindrical shell method is: V=(circumference)(height)(thickness) or (2pi*r)(r*h)(delta r). So I set up the integral as (2pi) integral [7x -x^5] dx. The boundaries are found by setting x^4=1, which yields -1 and 1. After differentiating we have 2pi[7/2x^2-1/6x^6] from -1 to 1. Because of my boundaries, I initially got the volume=0, but I don't think that's possible. I assumed the minus sign should be a plus, but after adding and multiplying by 2pi, I still got the wrong answer.

Any clues would be much appreciated! Thanks!
 
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  • #2
You're rotating the region about the line y=7 right?
Then if you're using cylindrical shells, you should integrate wtr y (from 0 to 1).
The radius of the shell is [itex]7-y^{1/4}[/itex].

Try setting up the integral again.
 
  • #3


Sure, I'd be happy to help. It looks like you have the right idea and set up the integral correctly. However, there are a few errors in your calculations.

First, when you set up the integral, it should be (2pi)(radius)(height)(thickness), so the function inside the integral should be (7-x^4) instead of (7x-x^5).

Second, when you integrate, you forgot to include the 2pi in front of the integral. So the integral should be 2pi∫[7-x^4]dx.

Third, when you plug in the boundaries, you should get 2pi[(7/2x^2)-(1/5x^5)] from -1 to 1. And when you solve this, you should get the correct answer of approximately 41.5 cubic units.

I hope this helps and clears up any confusion. Keep practicing and you'll get the hang of it again!
 

What is the Cylindrical Shell Method?

The Cylindrical Shell Method is a technique used in calculus to calculate the volume of a solid with a cylindrical shape. It involves breaking down the solid into thin cylindrical shells and integrating their volumes to find the total volume of the solid.

When is the Cylindrical Shell Method used?

The Cylindrical Shell Method is typically used when the shape of the solid cannot be easily described using traditional methods, such as the disk or washer method. It is also useful when the solid has a variable cross-sectional area along its height.

What are the steps involved in using the Cylindrical Shell Method?

The steps involved in using the Cylindrical Shell Method are as follows:
1. Visualize the solid and determine its boundaries
2. Determine the radius of the cylindrical shells
3. Set up the integral for the volume of one cylindrical shell
4. Integrate to find the total volume of all the shells
5. Simplify the integral and solve for the volume.

What are the advantages of using the Cylindrical Shell Method?

One advantage of using the Cylindrical Shell Method is that it allows for the calculation of volumes for solids with irregular shapes or varying cross-sectional areas. It also often requires fewer steps and can be more intuitive compared to other methods.

Are there any limitations to using the Cylindrical Shell Method?

The Cylindrical Shell Method may not be suitable for all types of solids. It is best suited for solids with cylindrical symmetry and may not work well for more complex shapes. Additionally, the calculation of the cylindrical shell radius can be challenging for some solids.

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