Volume of Rotated Region: y=x and y=x^2, about x-axis and y=2

[fk378]

Homework Statement

(a)The region R enclosed by the curves y=x and y=x^2 is rotated about the x-axis. Find the volume of the resulting solid.

(b)Find the volume of the solid in part (a) obtained by rotation the region about y=2.

The Attempt at a Solution

I solved the (a) integral and got 2(pi)/15, which is the right answer. I thought that (b) would give the same answer, since I'm just rotating it about a different line. But my answer (which is right) is 8pi/15. Can anyone explain why the volume would be different? It's the same solid, just rotated about different lines, right?

 

[Dick]

They are not 'the same solid'. The radius of rotation in the two different cases is different. Look up Pappus' theorem.

 

[badphysicist]

Let's take the example of making a donut in a similar manner. Take an oval centered on the origin where the longest radius is in the y-direction. Make a solid by rotating this region about the axis x=10 where 10 > the longer radius of the oval. This will make a donut and you could find the volume, but if you were to instead make a donut by rotating around y=10, it would not have the same volume as the solid made by rotating around x=10. I would expect that the y=10 solid would have a larger volume since it will have a greater distance from the center to the edge of the solid.

In your case, you're making some weird solids. and the distance from the rotational axis to farthest out point is not even the same in both cases. Around the x-axis, the farthest out point from the axis is x=1 so a magnitude of 1 away. For the y=2 axis, the farthest point away from it is x=0 for a magnitude of 2 away. I hope this helps you visualize this better.