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?

Question

Emilia Aguilar · Accepted Answer

Volume #= 8pi #

When #y(2-x)# is rotated about the #y#-axis it will generate a cone of base radius #2# and height #6#, hence the volume is:

# \ \ \ \ \ 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 #y#-axis is given by:

# 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 #

Scarlett Allison · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the graphs $ y = 3(2 - x) $, $ y = 0 $, and $ x = 0 $ about the y-axis, you can use the method of cylindrical shells.

1. Determine the limits of integration by finding the intersection points of the curves.
2. Set up the integral for the volume using the formula for the volume of a cylindrical shell.
3. Integrate the expression to find the volume.

The volume $ V $ can be calculated using the formula:

$$ V = \int_{a}^{b} 2\pi x f(x) \, dx $$

where $ f(x) $ is the height of the shell at $ x $, and $ a $ and $ b $ are the limits of integration.

In this case, the region is bounded by the y-axis and the curve $ y = 3(2 - x) $. The limits of integration are from $ x = 0 $ to $ x = 2 $, as those are the x-values where the curve intersects the y-axis.

So, the volume $ V $ is given by:

$$ V = \int_{0}^{2} 2\pi x \cdot (3(2 - x)) \, dx $$

You can evaluate this integral to find the volume of the solid.