Solids of revolution problem

jeepers101

Hi everyone,

I'm stuck on a problem. I have two approaches to the problem and am getting two different answers. Can you tell me which one is wrong and why.

Question:

y^2 = x, 2y = x; About the y-axis


Now first approach is to take (y^2 - 2y) and sub that into the SOR formula.

The second approach would be to find the volume of the two shapes separately and subtract them.

Thanks in advance.

TheBody

jeepers101 wrote: »

Hi everyone,

I'm stuck on a problem. I have two approaches to the problem and am getting two different answers. Can you tell me which one is wrong and why.

Question:

y^2 = x, 2y = x; About the y-axis


Now first approach is to take (y^2 - 2y) and sub that into the SOR formula.

The second approach would be to find the volume of the two shapes separately and subtract them.

Thanks in advance.

Hi there!

You should have gotten the same answer both ways. Maybe you could post the calculations you did so we could have a look at it to see where you went wrong. You could scan the page and post it here.

jeepers101

I can't scan the page at the moment but I can tell you where the difference is coming from. If I take the first approach I get my area to be the integral of (pie times)

(y^2 - 2y)^2

which becomes

y^4 - 4y^3 + 4y^2

The second approach yields a volume of

y^4 - 4y^2

You can see that the first approach has the term '-4y^3', which is not in the second.

TheBody

jeepers101 wrote: »

I can't scan the page at the moment but I can tell you where the difference is coming from. If I take the first approach I get my area to be the integral of (pie times)

(y^2 - 2y)^2

which becomes

y^4 - 4y^3 + 4y^2

The second approach yields a volume of

y^4 - 4y^2

You can see that the first approach has the term '-4y^3', which is not in the second.

Ah I see your mistake now!! Your error is in your first method. It's in the (y^2 - 2y)^2 bit. The formula has the ^2 on each individual part not on the whole piece. So to do it in one go you should have pi times the integral of (y^2)^2-(2y)^2. Effectivly the formula is just saying that the volume of revolution between the curves is just one volume minus the other one.

Hope that makes sense!!

jeepers101

Yes, that does make sense BUT...

I am under the impression that if we are looking for the area between the line and the curve we could use the first method

ie int(y^2 - 2y)

Now if this gives us the area between the line and the curve then surely subbing it into the SOR formula would give us the volume of this area as it rotates around the axis?

TheBody

I see what you are getting at. What you need to do is check out a calculus book to see how the formula comes about. It's kinda hard to explain here but I'll give it a go!!.....

The formula you are using is v=pi times integral (radius)^2 dx/dy. This is called the disk method in some books. Think about how the solid looks when revolved. If you were to slice it up into disks, and add up the volume of those disks, you would get the total volume.

To begin, the volume of a regular disk is V=(Pi)(radius)^2(w) where w=the width of the disk. In our case the disk is gonna be sliced up into infinitely skinny slices of width dx or dy (depending on which axis you are talking about). So we need to sum up all these infinitely skinny disks to get the volume. The way we do that is to ingetrate. So using the formula from above (V=(Pi)(radius)^2(w)) we see that

V=Pi times the integral of(radius)^2 dx/dy

where w gets replaced by (sort of!!) dx or dy .

Check out this youtube video which describes what I was trying to say above.

http://www.youtube.com/watch?v=1CbZlM09zF8&feature=related

jeepers101

Great link. Thanks.

TheBody

Glad I could help!!