Consider a circle with radius
centered at some point
on the
-axis. This circle has equation

Revolve the region bounded by this circle across the
-axis to get a torus. Using the shell method, the volume of the resulting torus is

where
.
So the volume is

Substitute

and the integral becomes

Notice that
is an odd function, so the integral over
is 0. This leaves us with

Write

so the volume is

Answer:
Give me any one dimension hieght or radius.
I just need to answer some questions lol
Subsitution is subsituting
y=3x-2
y=-x+6
subsitute 3x-2 for y in second equation
3x-2=-x+6
add x to both sides
4x-2=6
add 2 to both sides
4x=8
divide both sides by 4
x=2
subsitute
y=-x+6
y=-2+6
y=4
x=2
y=4