From what I understand, I am supposed to convert everything to terms of $y$, but I'm not sure how to solve for $x$ in $y=x\sqrt{2-x}$.
So far I have the setup of: $$2 \pi \int_0^{\frac{4 \sqrt{6}}{9}}(|y-(-5)|) \cdot(\text { new solved for } x \text { equation })\cdot dy$$
I used the function maximum $y$-value, $\frac{4 \sqrt{6}}{9}$, as the top of my interval.
Please help me figure out what I'm missing or doing wrong. Thank you!
If $f(x)=x\sqrt{2-x}$ then we want values from $x=0$ to $x=2.$
At any value $x,$ we have a cross section of a disk of radius $5+f(x)$ minus a disk of radius $5.$
So this volume is equal to:
$$\int_0^2\pi\left((f(x)+5)^2-5^2\right)\,dx$$
If you rotated around $x=-5,$ you would need to integrate by $y,$ and figuring out the value of the integral would be much harder, because for each $y$ you would have to find the two real $x_1<x_2\in[0,2]$ with $f(x_i)=y.$ Then you'd want:
$$\int_{0}^{y_{max}} \pi(x_2^2(y)-x_1^2(y))\,dy$$
But that is going to be much harder - finding formula for both $x_1,x_2$ involves solving a cubic.