How do you find volume by rotating area enclosed by #y=x^3# and #y=sqrt(x)# about x=1?

Question

Isaiah Allen · Accepted Answer

See the answer below:

Christopher Allers · Answer

undefined To find the volume by rotating the area enclosed by $y = x^3$ and $y = \sqrt{x}$ about $x = 1$, you would use the method of cylindrical shells. The formula for this method is:

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

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

First, find the intersection points of $y = x^3$ and $y = \sqrt{x}$ by setting them equal to each other and solving for $x$:

$$ x^3 = \sqrt{x} $$
$$ x^6 = x $$
$$ x^6 - x = 0 $$
$$ x(x^5 - 1) = 0 $$
$$ x = 0 \text{ or } x^5 = 1 $$
$$ x = 0 \text{ or } x = 1 $$

So, $x = 1$ is the only intersection point.

The integral becomes:

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

Now integrate to find the volume.