Check the picture below.
so by graphing those two, we get that little section in gray as you see there, now, x = 6 is a vertical line, so we'll have to put the equations in y-terms and this is a washer, so we'll use the washer method.

the way I get the radii is by using the "area under the curve" way, namely, I use it to get R² once and again to get r² and using each time the axis of rotation as one of my functions, in this case the axis of rotation will be f(x), and to get R² will use the "farthest from the axis of rotation" radius, and for r² the "closest to the axis of rotation".

now, both lines if do an equation on where they meet or where one equals the other, we'd get the values for y = 0 and y = 1, not surprisingly in the picture.
![\displaystyle\pi \int_0^1\left( 3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16} \right)dy\implies \pi \left( \left. \cfrac{3y^2}{2} \right]_0^1-\left. y^3\cfrac{}{} \right]_0^1-\left. \cfrac{y^3}{48}\right]_0^1+\left. \cfrac{y^5}{80} \right]_0^1 \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{59\pi }{120}~\hfill](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cpi%20%5Cint_0%5E1%5Cleft%28%203y-3y%5E2-%5Ccfrac%7By%5E2%7D%7B16%7D%2B%5Ccfrac%7By%5E4%7D%7B16%7D%20%5Cright%29dy%5Cimplies%20%5Cpi%20%5Cleft%28%20%5Cleft.%20%5Ccfrac%7B3y%5E2%7D%7B2%7D%20%5Cright%5D_0%5E1-%5Cleft.%20y%5E3%5Ccfrac%7B%7D%7B%7D%20%5Cright%5D_0%5E1-%5Cleft.%20%5Ccfrac%7By%5E3%7D%7B48%7D%5Cright%5D_0%5E1%2B%5Cleft.%20%5Ccfrac%7By%5E5%7D%7B80%7D%20%5Cright%5D_0%5E1%20%5Cright%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Ccfrac%7B59%5Cpi%20%7D%7B120%7D~%5Chfill)
D'(-3,-4)
E'(-3,-10)
F'(3,-6)
Answer:
4
Step-by-step explanation:
Answer:
Then maybe pay attention in class and know what to do and not ask strangers for help. We answer to try and help. Not like we get paid. We literally are just trying to help