Set Up an Integral Representing the Volume of the Torus $(x-6)^{2}+y^{2}=1$

Multiple Choice

Set up an integral representing the volume of the torus (donut) obtained by rotating the circle $(x-6) ^{2}+y^{2}=1$ about the y-axis.

A) $\int_{-1}^{1} \pi\left(6+\sqrt{1-y^{2}}\right) ^{2} d y$
B) $\int_{-1}^{1} \pi\left(6-\sqrt{1-y^{2}}\right) ^{2} d y$
C) $\int_{-1}^{1} \pi\left(\left(6+\sqrt{1-y^{2}}\right) ^{2}+\left(6-\sqrt{1-y^{2}}\right) ^{2}\right) d y$
D) $\int_{-1}^{1} \pi\left(\left(6+\sqrt{1-y^{2}}\right) ^{2}-\left(6-\sqrt{1-y^{2}}\right) ^{2}\right) d y$