The area bounded by the curve #y=3+2x-x^2# and line #y=3# is rotated completely about the line #y=3#. Find the volume of the solid of revolution obtained?

1 Answer
Jun 10, 2015

Answer:

#V=16/15pi~~3.35103#

Explanation:

The area are the solution of this system:
#{(y<=-x^2+2x+3),(y>=3):}#
And it is sketched in this plot:
enter image source here
The formula for the volume of a x-axis rotation solid is:
#V=pi*int_a^b f^2(z) dz#.

To apply the formula we should translate the half moon on the x-axis, the area won't change, and so it won't change also the volume:
#y=-x^2+2x+3color(red)(-3)=-x^2+2x#
#y=3color(red)(-3)=0#
In this way we obtain #f(z)=-z^2+2z#.
The translated area now is plotted here:
enter image source here
But which are the a and b of the integral? The solutions of the system:
#{(y=-x^2+2x),(y=0):}#
So #a=0 and b=2#.

Let's rewrite and solve the integral:
#V=pi*int_0^2 (-z^2+2z)^2 dz#
#V=pi*int_0^2 z^4-4z^3+4z^2 dz#
#V=pi*[z^5/5-(4z^4)/4+(4z^3)/3]_0^2#
#V=pi*[z^5/5-z^4+(4z^3)/3]_0^2#
#V=pi*(2^5/5-2^4+(4*2^3)/3-0^5/5+0^4-(4*0^3)/3)#
#V=pi*(32/5-16+32/3+0)#
#V=pi*(96/15-240/15+160/15)#
#V=pi*(96/15-240/15+160/15)#
#V=16/15pi~~3.35103#
And this "lemon" is the solid obtained:
enter image source here