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.
?
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
To find the volume of the solid obtained by rotating the region bounded by the given curves about the line x=1, you can use the method of cylindrical shells.
First, let's see how the region looks like. The region is bounded by the curves y=x, y=0, x=4, and x=6. When you sketch this region, you will see that it is a triangle with a base of length 2 (from x=4 to x=6) and a height of 4 (from y=0 to y=4).
To set up the integral for the volume using cylindrical shells, you need to express the radius and height of each shell in terms of y.
For each value of y, the radius of the shell will be the distance between the line x=1 and the curve y=x. So the radius will be 1 minus the value of y.
The height of each shell will be the difference in x-coordinates between the boundaries of the region at that y-value. In this case, the boundaries are x=4 and x=6. So the height will be 6 minus 4, which is 2.
Now, we can write the integral for the volume:
V = ∫(2π(1-y) * 2)dy, evaluated from y=0 to y=4
Simplifying, we have:
V = ∫(4π(1-y))dy, evaluated from y=0 to y=4
Integrating, we get:
V = 4π∫(1-y)dy, evaluated from y=0 to y=4
= 4π[(y - (y^2/2))] evaluated from y=0 to y=4
= 4π[(4 - (16/2)) - (0 - (0/2))]
= 4π[4 - 8]
= 4π*(-4)
= -16π
Therefore, the volume of the solid obtained by rotating the region about the line x=1 is -16π cubic units.