How do you find the volume of the solid generated by revolving the region bounded by the curves y=x^3 and y=x^4 rotated about the y-axis?

Question

Paisley Johnson · Accepted Answer

This is something that can be done two ways---the earlier way and the Shell Method. The Shell Method isn't too difficult to apply.

In this method, since you are rotating about the y-axis, your thickness #f(x)# is the function farther right minus the function farther left. If you get a negative volume at the end, you know you did it backwards.

Your graph basically looks like:

graph{(x^3 - y)(x^4 - y)sqrt(0.5^2 - (x-0.5)^2)/sqrt(0.5^2 - (x-0.5)^2) <= 0.00 [-1, 2, -0.095, 2]}

from #x = 0# to #x = 1#.

The formula uses the idea of circumference (the #2pix#) with the varying thickness #f(x)# of the solid along the vertical direction to build the solid by stacking shells vertically, while the radius #x# indicates the distance from the axis of rotation; since the axis of rotation is the y-axis, the radius is simply #x#, spanning #0# to #pm1#.

#V = int2pixf(x)dx#

#= 2piint x(x^3 - x^4)dx#

#= 2piint x^4 - x^5dx#

#= 2pi [1/5x^5 - 1/6x^6]|_(0)^(1)#

#= 2pi [(1/5 - 1/6) - 0]#

#= color(blue)(pi/15 "u"^3)#

Emily Ammerman · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the curves $y = x^3$ and $y = x^4$ rotated about the y-axis, you can use the method of cylindrical shells.

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

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

Where $f(x)$ and $g(x)$ are the functions representing the curves bounding the region, and $a$ and $b$ are the x-values where the curves intersect.

For $y = x^3$ and $y = x^4$, solving $x^3 = x^4$ gives $x = 1$, so $a = 0$ and $b = 1$.

Thus, the volume $V$ is given by:

$$ V = 2\pi \int_{0}^{1} x \cdot (x^4 - x^3) \, dx $$

You can then integrate this expression to find the volume.