What is the volume of the solid produced by revolving #f(x)=x^3, x in [0,3] #around #y=-1#?

1 Answer
Feb 7, 2018

See explanation below

Explanation:

We have a revolution solid with a hole inside because the spin axis is y=-1.
Consider a volume element with a width of #dx# and two radios:
#R= 1 + x^3# and #r=1#.
So, the volume of this volume element is #dV=pi(R^2-r^2)dx#
The sbstitution in that formula give us:

#dV=pi((1+x^3)^2-1^2)dx#

Integrating between 0 and 3 wil give us the volume requested

#int_0^3 pi(1+2x^3+x^6-1)dx=int_0^3pi(2x^3+x^6)dx#

solving this integral we have:

#V=2pix^4/4+x^7/7}_0^3=81pi/2+2187pi/7=4941/14pi#