Calculating Volume of Solid Using Cylindrical Shell Method

[pattiecake]

"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!

 

[Galileo]

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 $7-y^{1/4}$.

Try setting up the integral again.

 

[M98Ranger]

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!