Volume of Cylindrical Shells?

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.

y = x^4 , y = 0 , x = 1 ; about x=2

I've tried various formulae, but none of them seem to yield the correct answer. I know the correct answer, just not the solution.

So far, I am veering towards V = ∫2π(x-2)(x^4)dx [integrated from 1 to 2], but the final answer does not match with the answer given by the book.

Question

Volume of Cylindrical Shells?

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.

y = x^4 , y = 0 , x = 1 ; about x=2

I've tried various formulae, but none of them seem to yield the correct answer. I know the correct answer, just not the solution.

So far, I am veering towards V = ∫2π(x-2)(x^4)dx [integrated from 1 to 2], but the final answer does not match with the answer given by the book.

isaiah.feynman · Answer

Were you given the limits of integration or you solved for it?

agent_a · Answer

I solved for the limits of integration.

isaiah.feynman · Answer

And you're sure its correct?

agent_a · Answer

Nope. I'm only sure about the limits of integration, if the rotation was about the x-axis or y-axis. I know you have to account for the radius on how it is not solely x or y respectively, but rather, you have to subtract a certain number from the radius, because the rotation is done on a line. From what I understand, it should be x-2, because you rotate on x =2, but that doesn't seem to be the case.

isaiah.feynman · Answer

Its not x-2 I think the radius is 2.

agent_a · Answer

But if 2 is the radius, the cylinder would have to extend towards the y-axis, until one of its sides is flush with it. Initially, I was integrating from 0 to 2, but I realized that that would result in the radius' switch, completely. Because, under those circumstances, 2 would not be the radius anymore, and 1 would be the axis of rotation.

agent_a · Answer

Either that, or it is a larger cylinder (in the instance that 2 is the radius), but that cannot happen because the region of x =1 and x = 2 will be surpassed... not to mention, there are no points to intersect with y=x^4 thereafter, because y=x^4 only intersects x =2 at (2, 16).

agent_a · Answer

I know that you have to subtract 2 from x, because the cylinder does not cover the area from 0 to 1... That's how you account for the difference... at least, that's what I've picked-up, from scanning through various material.

isaiah.feynman · Answer

Tricky

agent_a · Answer

Thanks for trying, though.

isaiah.feynman · Answer

No problems.

isaiah.feynman · Answer

But I haven't stopped trying, when I get it I will tell you. Wait  what is the correct answer to this problem?

agent_a · Answer

$$\frac{ 7 }{ 15 } \pi $$

isaiah.feynman · Answer

Ok thanks.

zepdrix · Answer

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