Posted by **christina** on Wednesday, January 29, 2014 at 3:15pm.

Consider the solid obtained by rotating the region bounded by the following curves about the line x=1.

y=x,y=0,x=4,x=6

Find the volume

So it would be

pi (integral from 3 to 6) of ((1-y)^2 -(1-0)^2) right?

so then you integrate it and get

pi(Y^3/3-y^2) from 3 to 6.

?

- Calculus -
**Steve**, Wednesday, January 29, 2014 at 5:34pm
If you are going to integrate over y, the solid has two parts: a plain old cylinder of height 4 and thickness 2, and a variable-thickness shape of height 2.

So,

v = π(5^2-3^2)(4) + ∫[4,6] π(R^2-r^2) dy

where R=5 and r=x-1=y-1

v = 64π + π∫[4,6] 25-(y-1)^2 dy

= 64π + π∫[4,6] -y^2+2y+24 dy

= 64π + 52/3 π

= 244/3 π

I think shells are easier in this case. SO, since each shell has thickness dx, we have

v = ∫[4,6] 2πrh dx

where r = x-1 and h = y = x

v = 2π∫[4,6] (x-1)(x) dx

= 2π∫[4,6] x^2-x dx

= 244/3 π

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