Find the volume when the plane area bounded by y=-x^2-3x+6 and x+y-3=0 is revolved about (1) x=3 and (2) about y=o

B. Find the volu6 of focus generated by revolving the circule x^2+y^2=4 about the line x=3

3 answers

A: The curves intersect at (-3,6) and (1,2)

So, consider the volume as a set of nested cylinders of thickness dx. Then

v = ∫[-3,1] 2πrh dx
where r=3-x and h=(-x^2-3x-6)-(3-x)
v = ∫[-3,1] 2π(3-x)((-x^2-3x-6)-(3-x)) dx
= 2π∫[-3,1] (3-x)(-x^2-2x+3) dx
= 256π/3

B. Google volume of torus

By integration