How can I find the volume generated by revolving the following region about $x=5$?

Question

The region enclosed by : $y=6-x^2$ and $y=5$ I first get the inverse functions and the intersections and then work with the disk/washer method, the result is zero and I can't figure out what am I doing wrong, I can still solve it using the shell method. I can't upload a picture for the graph /my work(I get an error from the app)

Answer 1

If $y= 6- x^2$ then $x= \pm\sqrt{6- y}$. For any y between 5 and 6, we have a "washer" with inner radius $5- \sqrt{6- y}$ and outer radius $5+ \sqrt{6- y}$ so with area $\pi(5+ \sqrt{6- y})^2- \pi(5- \sqrt{6- y})^2= \pi(25+ 10\sqrt{6- y}+ 6- y)-\pi(25- 10\sqrt{6- y}+ 6- y)$$= 20\pi\sqrt{6- y}$.

Integrate that, with respect to y, from y= 5 to y= 6.