# Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.

Find the volume generated by rotating the region bounded by the given curves about the specified axis. The curve is given by,

$$y = x^3$$

with bounds $$y=8$$ and $$x=0$$, and about the rotating axis $$x=3$$.

Here is my current attempt:

$$V= 2\pi \int_0^8 (x-3)x^3=2\pi \int_0^8 (x^4 - 3x^3) dx$$

$$= 2\pi \left( \frac{x^5}{5} -\frac{3x^4}{4} \right)_0^8$$

The answer I get is a ridiculous number. I am unsure on how to get the limits of integration from these question in general so that makes it more challenging.

Question: 1. How do I find the limits of integration in questions such as these? 2. What am I doing wrong in this question?

• Just click on "Edit" to see the MathJax code that corresponds to the equations. For reference see MathJax quick guide. Commented Sep 15, 2019 at 14:43

Let $$u= x-3$$ and the curve becomes

$$y=(u+3)^3$$

So, the volume is equivalent to that of rotating the above curve around $$u=0$$. Since $$y(-1) = 8$$, the bounds for integration is $$(-3,-1)$$. The corresponding integral is

$$V = \int_{-3}^{-1} 2\pi |u|(8-y)du$$ $$=-\int_{-3}^{-1} 2\pi u[8-(u+3)^3]du=\frac{264}{5}\pi$$

Of course, you could integrate without the above variable change. Then, the correct integral expression should be,

$$V = \int_{0}^{2} 2\pi (3-x)(8-x^3)dx=\frac{264}{5}\pi$$

Note the upper bound is $$x=2$$ since the volume is capped at $$y=x^3=8$$ and the volume is between $$y=8$$ and $$y=x^3$$, hence $$8-x^3$$ in the integrand.

• First of, thank you for changing my numbers and letters into an equation in my question. How did you do that if you don't mind me asking? Commented Sep 14, 2019 at 20:06
I'm thinking you should integrate from $$x=0$$ to $$x=2$$. Drawing a picture should help.
So $$2\pi\int_0^2(3-x)(8-x^3)\operatorname dx=2\pi\int_0^2(24-8x-3x^3+x^4)\operatorname dx=2\pi[24x-4x^2-\frac{3x^4}4+\frac{x^5}5]_0^2=2\pi(48-16-12+\frac{32}5)=\frac{264\pi}5$$.
• It should be $8-x^3$. I made a mistake. Wrong region.