Is An Infinitely Thin Cylindrical Shell a Rectangle?

Question

Yesterday I finished reading the method for finding the volume of a solid of revolution using cylindrical shells, the textbook I use of course gave a rigorous proof on why it works, however, it also showed a neat method to think about it; that is, to cut open the cylindrical shell of width, say chance $\Delta x$, into an somewhat of a rectangle. The textbook says that the reader should observe that the volume of the cylindrical shell is approximately that of the rectangle and the diagram could be used to remember the formula. However, wouldn't the volume of the cylindrical shell be exactly that of the rectangle when it's width, $\Delta x$, becomes infinitesimally small? (That is, $dx$). For then the outer radius would be equal to the inner radius and coalesce into one radius forming exactly one circumference. So when it's cut open the circumference should exactly coincide with the length of the rectangle. I ask because the textbook hasn't clearly stated this. Is this so? Or am I commiting an error in reasoning? Thank you in advance.

Answer 1

There is no such thing as infinitesimally small. No matter how small $\Delta r$ gets the outer circumference will be $2\pi r+2\pi\Delta r$ where $r$ is the inner radius. It is in the limit as $\Delta r\to 0$ that the two become equal.