Volume of 3D objects

Question

The graph is to be revolved around the line $y=4$

Consider the functions $$f(x)=\frac{2}{x+1}$$ $$y=0$$ $$x=0$$ $$x=4$$ So this means we need to use the washer method but where i'm stuck on is what $r$ is meant to be. I know $R$ is $R(x)=4- \frac{2}{x+1}$ but should I use $x=0$ and $x=4$ or just one of them for $r$?

Answer 1

Washer method is fine.

The larger radius is $4$ and the smaller radius is

$$4-\frac {2}{x+1}$$

The bounds of your integral are $x=0$ and $x=4$

The integrand is $$\pi (4^2-( 4-\frac {2}{x+1})^2)$$

Answer 2

HINT

By disk method, the set up should be

$$V=\int_0^4 \pi 4^2dx-\int_0^4 \pi[4-f(x)]^2 dx$$

Answer 3

You have to use every element 0 to 4 since you are integrating, so Your volume is $$V=\int_{0}^{4}\pi[4-{2 \over x+1}]^2dx$$