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

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The Answer of 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) $\int_{-1}^{1} \pi\left(\left(6+\sqrt{1-y^{2}}\right)^{2}-\left(6-\sqrt{1-y^{2}}\right)^{2}\right) d y$

Set Up an Integral Representing the Volume of the Torus (x−6)2+y2=1(x-6)^{2}+y^{2}=1(x−6)2+y2=1

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