how to solve using washer method? or the circular ring method?

Question

isabel☺ · Answer

@ZeHanz ? help?

zehanz · Answer

Washer method: 
Suppose you have two graphs form x=a to x=b.
The area between the graphs is being rotated around the x-axis.
The volume swept out is then:$$V=\pi \int\limits_{a}^{b}(f^2(x)-g^2(x))dx$$It is called the washer method, because you can consider the volume as an infinite number of slices (washers) with an outer radius of f(x), an inner radius of g(x) and width dx.

zehanz · Answer

Circular ring method:
Often used to calculate the volume of a body you get by rotating the graph of a function y=f(x) around the y-axis.
The body is considered to be an infinite number of hollow cylinders, with radius x, height f(x) and thickness dx. The volume of one cylinder is 2pi x f(x) dx, so the total volume is given by$$V=\int\limits_{a}^{b}2\pi x f(x)dx$$

zehanz · Answer

Typo: The second method I described is often called the cylindrical shell method, or just shell method. I don't know if you meant to ask about it. 

The shell method is less well known than the first one, but it often leads to simpler integrands, because there is no squaring involved.

zehanz · Answer

I will give you an example of the first method.
It is easier if there is only one graph to be rotated around the x-axis:

$V=\pi \int\limits_{a}^{b}f^2(x)dx$    (you can consider g(x)=0 for every x).

Suppose $f(x)=\frac{1}{4}x^2$, and we want to rotate the area between x=2 and x=4 around the x-axis. See image for the graph.

The volume is then: $V=\pi\int\limits_{2}^{4}(\frac{1}{4}x^2)^2dx=\pi\int\limits_{2}^{4}\frac{1}{16}x^4dx=\frac{\pi}{16}\int\limits_{2}^{4}x^4dx$.

So $V=\frac{\pi}{16}\cdot \frac{1}{5}[x^5]_{2}^{4}=\frac{\pi}{80}(1024-32)=\frac{992\pi}{80}=\frac{62\pi}{5}$.

isabel☺ · Answer

why g(x) is 0?

isabel☺ · Answer

@ZeHanz

zehanz · Answer

Because there is only the graph of f that is being rotated  around the x-axis.

isabel☺ · Answer

oh  .. :) thanks!

zehanz · Answer

YW!