volumes of revolution

the region enclosed by the curve y=|2x-4| and the coord axes is rotated about y axis to form a solid, S. find the volume of S.

What do you do about the mod?

My thoughts are that when x<2, the graph is negative. So if you integrate pi((y-4)/-2)^2 between 0 and 4 (rather than pi((y+4)^2/4) dy). Is this the correct approach?

From a sketch of the curve you're interesteed in the x values from 0 to 2, and as you noticed, in this region what's in the mod is negative.

So, for this region y=4-2x

And thus x=(4-y)/2

And you have $\displaystyle\pi \int^4_0 [(4-y)/2]^2\; dy$

thanks! Do you know how to help with this one?:

The points P(3,2) and Q(0,1) lie on y^2 = x + 1
a) volume of solid generated when region bounded by y=0, x=0 and x=3 and the arc PQ of the curve is rotated about x axis. COMPLETED: 15pi/2

b)S is the region bounded by y=1, x=3 and the arc PQ of the curve. Show that when S is rotated completely about the x-axis the volume of the solid generated is 9pi/2.

Not sure how to do part b?