Volume of a Solid of Revolution?

Question

Fairly quick question, but I'm a bit confused on whether or not the work I did was in any way right. So the question went something like this: "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 $\int^7_0(2^2-(\sqrt[3]{y})^2)$ 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 $x=6$.

Answer 1

I believe the best approach to solving these types of problems is sketching out the region in question. So, we are dealing with the following region

Now, we can go about finding the volume. Since we are revolving around the y-axis, we can use the following formula: $$ V = \int_a^b A(y) dy$$

And, we know that $A(y) = \pi((\text{outer radius})^2 - (\text{inner radius})^2)$, 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 $y^{1/3} - 0 = y^{1/3}$.

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: $$ A(y) = \pi( (y^{1/3})^2 - (0)^2 ) = \pi(y^{2/3}) $$

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: $$ V = \int_0^8 \pi(y^{2/3}) dy $$

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.

Answer 2

Note that $y=x^3, \>x=0$ and $y=8$ only specify the area to be rotated. The volume is then defined by rotating this area around a given axis.

For instance, if the axis of rotation is $x=6$, the volume can be integrated with the disk method

$$\int_0^8 \pi[6^2-(6-\sqrt[3]y)^2]dy$$