Integrals: Volume of rotating the area bounded by 6x and x^2+3x about the x axis

Question

hitaro9 · Answer

What should the integral even look like here?

hitaro9 · Answer

I got the integral from 0 to 3 of pi ( 3x-x^2)^2 dx 

But webassign says thats wrong, so I'm not really sure.

accessdenied · Answer

Do you know what the region looks like / where it stands relative to the x-axis?
It is sort of like having the volume of rotating the top figure about the x-axis, and removing the volume contribution of the curve revolved about the x-axis underneath it.

hitaro9 · Answer

I imagine it would look something like this: |dw:1396207595609:dw|

Kinda like a cone with the middle part taken out

hitaro9 · Answer

With the tip of the cone being the origin?

accessdenied · Answer

It seems you subtracted the two values to find your radius, but remember that the distance to the x-axis is important when we revolve around the x-axis.  The same space of a radius directly adjacent to the x-axis will create a much larger area when it is further away:
|dw:1396207605170:dw|