SOLUTION: If the area of the region bounded by the graph of y=x^3, the line x=2 and the x-axis is revolved about the line x=-1 then find the volume of the solid revolution.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: If the area of the region bounded by the graph of y=x^3, the line x=2 and the x-axis is revolved about the line x=-1 then find the volume of the solid revolution.       Log On


   

Question 1101661: If the area of the region bounded by the graph of y=x^3, the line x=2 and the x-axis is revolved about the line x=-1 then find the volume of the solid revolution.
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If the area of the region bounded by the graph of y=x^3, the line x=2 and the x-axis is revolved about the line x=-1 then find the volume of the solid revolution.
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The limits of x are 0 and 2.
Use shells:
Volume of a shell at x = dx*x^3*2pi*(x+1)
dV = 2pi*(x^4 + x^3)dx
Vol = 2pi*(x^5/5 + x^4/4)
0 --> 0
For x = 2:
Vol = 2pi*(32/5 + 4)
= 104pi/5 cubic units
=~ 65.345 cubic units

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


I was hoping another tutor would answer this question, so that I could try to solve it and compare answers. I often have trouble setting up this kind of problem correctly, so I was glad to see another solution.

I did come up with the same answer as the other tutor, but by a different method.

So you can look at the two methods and see if you have a preference for one or the other.

The other tutor used the shell method in his solution. I generally have better luck with the disk method; so that is what I used.

And so that I could visualize the problem better, I found the answer by finding the volume as the volume of a cylinder, minus the volume of the solid of revolution between the curve and the line y=8, instead of finding the volume of the solid of revolution between the curve and the line y=0 (i.e., the x-axis).

So I'm integrating in the y direction from 0 to 8; and my disk has volume pi%2Ar%5E2%2Ady.

The curve is y+=+x%5E3; since I'm integrating in the y direction, I convert this to x+=+y%5E%281%2F3%29

Since the rotation is about the line x = -1, the radius r of my disk is y%5E%281%2F3%29%2B1

So the volume of my disk is pi%2A%28y%5E%281%2F3%29%2B1%29%5E2%2Ady or pi%2A%28y%5E%282%2F3%29%2B2y%5E%281%2F3%29%2B1%29%2Ady.

Doing the integration, the volume of my solid of revolution is then pi times %283%2F5%29y%5E%285%2F3%29%2B%283%2F2%29y%5E%284%2F3%29%2By evaluated between 0 and 8.

For y=8: %283%2F5%29%2A32%2B%283%2F2%29%2A16+%2B+8+=+256%2F5
and of course 0 for y=0.

The volume of my cylinder is pi%283%5E2%29%288%29+=+72%2Api

So the volume we are looking for in the problem is 72%2Api+-+%28256%2F5%29%2Api+=+%28104%2F5%29%2Api

... which is the answer the other tutor got.