Find the volume of the solid generated by revolving the region bounded by the above graphs of the equations around the line y=6.
My Attempt $\int_1^6(6- 6/x)) ^2 dx$
Why would we need two integrals? The region is consistently defined by the curve y=6/x. My teacher wrote two integrals and summed them. She wrote $\int_1^3 (6-6/x))^2dx + \int_3^6 16 dx$
16 comes from $\pi r^2$.
She assumes the second region has a constant radius but that isn't true.
For $1\le x\le3$, the radius $R=6-6/x$, but for $3\le x\le6$, the radius $R=4$; so
the volume of the solid is given by $\displaystyle V=\int_1^3\pi\big(6-\frac{6}{x}\big)^2dx+\int_3^6\pi(4)^2 dx$.
More simply, using the shell method gives $\displaystyle V=\int_2^62\pi(6-y)\big(6-\frac{6}{y}\big) dy$.