Using Disc Method to solve Integrand

Question

So I'm trying to figure the following out:

I'm just trying to set it up correctly to use on my graphing calculator and I got confused by something.

I'm aware of the formula for solving this:

$\int\pi (f(x))^2\,\mathrm{d}x$

I know that the integration limits are $x=0$ and $x=2$, what I'm trying to understand is, why is this the correct way of solving this?:

$\pi\int_0^2\left(\frac {2e^x}{1+e^x}\right)^2\,\mathrm{d}x-\int^2_0 (\pi)\,\mathrm{d}x$

I mean, the first integrand is obvious, yet what about the second one? The exercise asks "enclosed by $f$ and the lines $x=2$ and $y=1$. Isn't $y=1$ and $x=2$ a part of $f$? Why do I have to create a new function. This really confuses me, because I really need to recognize when I need to create a second integrand.

Thanks in advance for any help!

Answer 1

If you were to just do $\pi\int_0^2\left(\frac{2e^x}{1+e^x}\right)^2\,\mathrm{d}x$ you would not be doing the volume bound by this function and $y=1$. Instead you are finding the volume of the function bound by the $x$-axis is revolved, you have to subtract in order to get the desired volume

Answer 2

The formula for finding the volume of revolution is $$\int_{a}^{b}\pi(R^2-r^2)dx$$

Where R=Outer Radius and r=Inner Radius

Always remember that When revolution is about x-axis, put functions in terms of 'x' (Solve for 'y')

When revolution is about y-axis, put functions in terms of 'y'(Solve for 'x')

In your problem you need to find the revolution in term around $x$-axis. So, the volume is $$\int_0^2\pi\left(\left(\frac {2e^x}{1+e^x}\right)^2-(1)^2\right) dx $$