Calculus
posted by christina .
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 ((1y)^2 (10)^2) right?
so then you integrate it and get
pi(Y^3/3y^2) from 3 to 6.
?

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 variablethickness shape of height 2.
So,
v = π(5^23^2)(4) + ∫[4,6] π(R^2r^2) dy
where R=5 and r=x1=y1
v = 64π + π∫[4,6] 25(y1)^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 = x1 and h = y = x
v = 2π∫[4,6] (x1)(x) dx
= 2π∫[4,6] x^2x dx
= 244/3 π