Question: The volume obtained by rotating the region bounded by $y=\sqrt{81-(x-4)^2}$ and the $x\ -$ axis about the line $x=14$.
I used both Disk and cylindrical shells method to evaluate the volume. Using Cylindrical Shells : Here the expression will be $V=\int_{-5}^{13} 2\pi\ (14-x)\sqrt{81-(x-4)^2} \ dx=810\pi^2$. Using Disk mehod : we have $\int_{0}^{9} \pi\ ((-10-\sqrt{81-y^2})^2-(\sqrt{81-y^2}-10)^2)\ dx= \int_{0}^{9}40 \pi \sqrt{81-y^2} \ dy=810 \pi^2$.
The answer is $405 \pi$. Please correct me. Thanks.
Where are you getting the $\pi^2$?
Notice that the shape is half donut since y cannot be negative. Also, in disk method there should be dy in the first integration. The shape for your reference: