Volumes of Solid of Revolution

[Flaneuse]

I have a volume problem that I has been bothering me for a while now, as I have just not been able to figure it out.

The question involves finding the volume of the solid generated by revolving the region bounded by y = x, y = 0, and y = 4 around the line x = 6. I tried doing $$\int$$
$$^{4}_{0}$$16 - (4-y)$$^{2}$$ dy + $$\int$$$$^{4}_{0}$$36-16 dy but that did not work out. I am fairly sure that the part of it that is giving me problems is the little extra area between x = 6 and x = 4, as the y = x and x =6 do not intersect due to the y = 4 boundary...

There is also another question like this that has been giving me problems, involving fining the volume of the solid generated by x = y^2 and x = 4 around the line x = 6; would this be done the same way?

Thanks!

 

[Dustinsfl]

By what you wrote, I would think the solid is being revolved around x=6; therefore, you need to use the formula V= pi* the integral of [R(x)]^2-[r(x)]^2 and little r is the distance from the solid to the axis of revolution. So the shape meets at (4,4) as the end point with a difference of 2 units to x=6.

I hope this helps.

 

[Flaneuse]

Oops, I seem to have left out the pis in the original post...

So, now I tried using (6-x) for R(x) and 2 for r(x) and integrating from 0 to 6, but I end up with 48 pi... The answer is supposed to be 208(pi)/3... What am I doing wrong?

 

[Dustinsfl]

Which one is 208 pi/3? The first or second? Also, the shell method maybe easier to use.

 

[Dustinsfl]

Maple is telling me the answer is (224/3)*pi

 

[Flaneuse]

The first one is supposed to be 208(pi)/3, or at least that is what it says in my textbook.

 

[Mark44]

I have a volume problem that I has been bothering me for a while now, as I have just not been able to figure it out.

The question involves finding the volume of the solid generated by revolving the region bounded by y = x, y = 0, and y = 4 around the line x = 6. I tried doing $$\int$$
$$^{4}_{0}$$16 - (4-y)$$^{2}$$ dy + $$\int$$$$^{4}_{0}$$36-16 dy but that did not work out. I am fairly sure that the part of it that is giving me problems is the little extra area between x = 6 and x = 4, as the y = x and x =6 do not intersect due to the y = 4 boundary...

There is also another question like this that has been giving me problems, involving fining the volume of the solid generated by x = y^2 and x = 4 around the line x = 6; would this be done the same way?

Thanks!

I don't believe you have posted the problem correctly. What you have described is the region between two parallel lines (y = 0 and y = 4), cut somehow by the line y = x. This does not determine a finite region.

Also, for future reference, use one pair of [ tex] and [ /tex] tags for the entire integral. You are putting way too many pairs of them, which results in a garbled-up mess. Here is your first integral (which probably has nothing to do with this problem):

$$\int_{0}^{4} 16 - (4-y)^{2} dy$$

Double-click the integral to see the LaTeX script that generates this integral.

 

[Flaneuse]

Well, I have rechecked the problem's statement in my textbook, and that is exactly what it says. I was assuming though that it meant for me to revolve a triangular area (bounded by y = x, y = 0, y= 4, and x=6) around x = 6.

And thank you for the help with the typing of the integrals :]. Goodness knows I needed it.

 

[Mark44]

In that case, I think your book has a typo. The region bounded by y = x, x = 0, and y = 4 is probably what they meant.

Probably the most natural way to do this is to use circular washers, disks with holes in them. The volume of a typical volume element is
$$\Delta V = \pi (R^2 - r^2) \Delta x$$

Here, R is the radius from the axis of rotation (the line y = 6) to the line y = x, and r is the radius from the axis of rotation to the line y = 4.