Dean Summers

2022-07-17

Volume of a Solid of Revolution?
Find the volume if the region enclosing $y={x}^{3},x=0,$, and $y=8$ is rotated about the given line. So the axis of rotation that was given was the y axis, but I'm a bit confused on what the given bounds of $y=8$ and $x=0$ have to do with it, but I pretty much ignored them and put the given equation in terms of y and used the integral ${\int }_{0}^{7}\left({2}^{2}-\left(\sqrt[3]{y}{\right)}^{2}\right)$ to solve, is this at all right? Any help would be appreciated Also looking for some guidance on how I'd go about finding the volume if the axis of rotation was something like $x=6$.

Rihanna Robles

Expert

Step 1
Note that $y={x}^{3},\phantom{\rule{mediummathspace}{0ex}}x=0$ and $y=8$ only specify the area to be rotated. The volume is then defined by rotating this area around a given axis.
Step 2
For instance, if the axis of rotation is $x=6$, the volume can be integrated with the disk method ${\int }_{0}^{8}\pi \left[{6}^{2}-\left(6-\sqrt[3]{y}{\right)}^{2}\right]dy$

Glenn Hopkins

Expert

Step 1
Now, we can go about finding the volume. Since we are revolving around the y-axis, we can use the following formula: $V={\int }_{a}^{b}A\left(y\right)dy$
And, we know that $A\left(y\right)=\pi \left(\left(\text{outer radius}{\right)}^{2}-\left(\text{inner radius}{\right)}^{2}\right)$, where the radius is just the distance from the curve to the axis of rotation.
Since the axis of rotation is the y-axis (equivalently, $x=0$), we know that are distances must be x-distances, so we must solve our y-functions in terms of x.
Our outer radius is just the distance from the curve $y={x}^{3}$ to the y-axis, we which is given by ${y}^{1/3}-0={y}^{1/3}$.
Step 2
Our inner radius is the distance from the y-axis to itself, which is 0.
Therefore, the expression for the area in terms of y is:
$A\left(y\right)=\pi \left(\left({y}^{1/3}{\right)}^{2}-\left(0{\right)}^{2}\right)=\pi \left({y}^{2/3}\right)$
Now, we just need to plug back into the equation for the volume, and figure out the bounds. Since this is a dy integral, we need to know the y-bound. From the region, we can see that the integral should go from 0 to 8, since the region is bounded by $y=0$ and $y=8$
Therefore, our integral becomes: $V={\int }_{0}^{8}\pi \left({y}^{2/3}\right)dy$
If the axis of rotation changes, you just need to change your expression for the area. For example, for an axis of rotation of $x=6$, you can keep the area in terms of y, but now your inner and outer radii expressions will be different.

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