My basic question is: When we think of the area under the graph and extending it in the 3 dimensions, we actually get a cylinder with height = $f(x)$, thickness = $dx$ and inner radius = $x$. Then the volume should be $\int \pi (2x+dx) \ dx\ f(x) $. Right? Then why is it: $\int \pi\ x f(x) dx$ ??
Finding the volume when a region is rotated about the $y$-axis?
0
$\begingroup$
calculus
graphing-functions
algebraic-curves
rotations
1 Answers
1
You just need to think carefully about what is going on with the method of shells. I suggest reading here (scroll down about halfway where they start talking about shells).
Another great resource that explains what you need is video #29 in this series. (The entire set of videos is excellent, with great visuals as well as step-by-step explanations.)
Most generally (and as the links above will explain), the method of shells produces an integral of the form $\int_a^b \pi\cdot(\text{shell radius})\cdot(\text{shell height})\,dx,$ i.e., $\int_a^b \pi\,x f(x)\,dx$ as you mentioned.