I am asked to find the maximum and minimum volumes when rotating the region between the graphs of $y=\sqrt{4-x}$ and $y=c(0\le c\le2)$ about the x-axis on the interval $[0,4]$. I want to use the washer method for this.
There are two possibilities, I think. If the line is above the radical and if the line is below the radical. I'm at a loss as to how to set this up.
$$\pi\int_{0}^{4}(c^{2}+x-4)dx=\pi\bigg[(c^{2}x+\dfrac{x^{2}}{2}-4x)\bigg]_{0}^{4}=\pi(4c^{2}-8)$$
If I set that equal to zero, I get $c=\sqrt{2}$. I know that is the answer for a minimum value. Mathematically, I don't understand how I am finding the minimum. I didn't take a derivative.
I also don't know how to go about finding the maximum values (they are $c=0,2$). If I set up the integrand as $4-x-c^{2}$ then I end up getting the same answer of $c=\sqrt{2}$.