The region bounded by?

Question

The region bounded by x^2-y^2=3 and x=2 is rotated about the line y=3 .
Using cylindrical shells, set up an integral for the volume of the resulting solid.

i think this is the right setup
ʃ2pi(3-y)(sqrt(3+y^2)


i think this is the right setup but im not sure what the limits of integration are.

Answer 1

I generally solve this sort of problem by Runge-Kutta.

Answer 2

Integral lower limit y = -1 to y = +1

2pi [3-y] [2 - sqrt(3 + y^2)] dy

(following the pattern of "thin" cylindrical shells abt y = 3, the radius of shell(s) in first bracket-pair; the variable length of shell(s) in second bracket pair; times a "thinness" dy make a mini- volume to be "added up" or integrated)

After doing the bracket multiplications there are 4 terms to integrate; 2 will be easy; the remaining 2 (I had to use trig substitution on) are loaded with opportunities for careless mistakes; one of those 2 is zero! My final answer

1.0562 x 2 pi = about 6.636