# 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 #