# Volumes of Disks and Washers

1. Apr 28, 2008

### zcabral

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by
y=x^2
y=2x
about the x-axis.
So i used intergral 0 to 2 (2x-x^2)^2 dx
this is wat i got....
pi*(((2^5)/5)-(4(2^4)/4)+(4(2^3)/3))
its wrong so wat did i do wrong?

2. Apr 28, 2008

### Feldoh

That region will make a washer. Your interval [0,2] looks right.

However it might be beneficial to find a small piece of volume of the actual shape.

V=A*h so an infinitely small piece of volume would be dV = A*dh, or more approately in our case:

dV=A*dx

The area is pi*r^2 since it forms a washer, so $$A=\pi (r_1^2-r_2^2)$$

Where r1 and r2 are the distances from the x-axis where r1>r2

The Area portion is what you did wrong, do you see where to go from there?

Next time you should post questions like this (non-conceptual) questions in the homework help.

Last edited: Apr 28, 2008
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