How do you find the volume of the solid generated by revolving the region bounded by the graphs #y=3(2-x), y=0, x=0#, about the y axis?
1 Answer
Dec 30, 2016
Volume
Explanation:
When
# \ \ \ \ \ V =1/3pir^2h #
# :. V = 1/3pi*2^2*6 #
# :. V = 8pi #
If you want a calculus solution, then the Volume of revolution about the
# V =int_a^b pi x^2dy#
Now:
# y=3(2-x) => y=6-3x #
# :. x=1/3(6-y) #
And so:
# V = int_0^6 pi (1/3(6-y))^2 \ dy #
# \ \ \ = 1/9 pi int_0^6 (6-y)^2 \ dy #
# \ \ \ = 1/9 pi int_0^6 (36-12y+y^2) \ dy #
# \ \ \ = 1/9 pi [36y-6y^2+1/3y^3]_0^6 #
# \ \ \ = 1/9 pi {(36*6-6*36+216/3) - (0)} #
# \ \ \ = 1/9 pi (72) #
# \ \ \ = 8 pi #