Calculate the volume $V$ of the solid $G$, which is obtained when from the sphere $x^{2}+y^{2}+z^{2}<36$ is cut by the cylinder $x^{2}+y^{2}\leq 1$. $G$ is given as: $$G:=\{(x,y,z), x^{2}+y^{2}+z^{2}<36, x^{2}+y^{2}\leq 1\}.$$
My solution: $$V=\int_{-\sqrt{35} }^{\sqrt{35} } \! \int_{0}^{2\pi} \! \int_{1}^{\sqrt{36-z^{2}} } \! r \, dr \, d\theta \, dz = \frac{140}{3}\sqrt{35}\pi.$$
Can someone tell me if it is correct?
Since you are cutting away the full cylinder $x^{2}+y^{2}\leq 1$ from the sphere $x^{2}+y^{2}+z^{2}<36$, I guess the solid $G$ is given by $$\{(x,y,z), x^{2}+y^{2}+z^{2}<36, x^{2}+y^{2}> 1\}.$$ In this case, your evaluation by using the cross-section method is correct.
Note that we can evaluate $V$ also by using the shadow method, $$V=\int_{r=1 }^{6} \! \int_{0}^{2\pi} \! \int_{z=-\sqrt{36-r^{2}}}^{\sqrt{36-r^{2}} } 1 dz\,(r \, dr \, d\theta)=4\pi\int_{r=1 }^{6} \sqrt{36-r^{2}} rdr\\ =\frac{4\pi}{3}\left[-(36-r^2)^{\frac{3}{2}}\right]_1^6 =\frac{140}{3}\sqrt{35}\pi.$$