How do you find the volume of the torus formed by revolving (x-2)^2 +y^2=1 about the y-axis?

Oct 18, 2015

$4 {\pi}^{2}$

Explanation:

${\left(x - 2\right)}^{2} + {y}^{2} = 1$ is a circle which center is $C \left(2 , 0\right)$ and radius $r = 1$.

$V = \pi {\int}_{{y}_{1}}^{{y}_{2}} \left({R}^{2} - {r}^{2}\right) \mathrm{dy}$

Obvious, ${y}_{1} = - 1 , {y}_{2} = 1$.

${\left(x - 2\right)}^{2} + {y}^{2} = 1 \implies x = 2 + \sqrt{1 - {y}^{2}} \vee x = 2 - \sqrt{1 - {y}^{2}}$

$R = 2 + \sqrt{1 - {y}^{2}}$
$r = 2 - \sqrt{1 - {y}^{2}}$

$V = \pi {\int}_{- 1}^{1} \left({\left(2 + \sqrt{1 - {y}^{2}}\right)}^{2} - {\left(2 - \sqrt{1 - {y}^{2}}\right)}^{2}\right) \mathrm{dy}$

V=pi int_(-1)^1 ((4+4sqrt(1-y^2)+1-y^2)-(4-4sqrt(1-y^2)+1-y^2)dy

$V = \pi {\int}_{- 1}^{1} \left(4 + 4 \sqrt{1 - {y}^{2}} + 1 - {y}^{2} - 4 + 4 \sqrt{1 - {y}^{2}} - 1 + {y}^{2}\right) \mathrm{dy}$

$V = \pi {\int}_{- 1}^{1} \left(4 + 4 \sqrt{1 - {y}^{2}} + 1 - {y}^{2} - 4 + 4 \sqrt{1 - {y}^{2}} - 1 + {y}^{2}\right) \mathrm{dy}$

$V = 8 \pi {\int}_{- 1}^{1} \sqrt{1 - {y}^{2}} \mathrm{dy}$

$y = \sin t \implies \mathrm{dy} = \cos t \mathrm{dt}$
${t}_{1} = \arcsin \left(- 1\right) = - \frac{\pi}{2}$
${t}_{2} = \arcsin 1 = \frac{\pi}{2}$
$\sqrt{1 - {y}^{2}} = \sqrt{1 - {\sin}^{2} t} = \sqrt{{\cos}^{2} t} = \cos t$

$V = 8 \pi {\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} {\cos}^{2} t \mathrm{dt} = 4 \pi {\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \left(1 + \cos 2 t\right) \mathrm{dt}$

$V = 4 \pi \left(t + \frac{1}{2} \sin 2 t\right) {|}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} = 4 \pi \cdot \pi = 4 {\pi}^{2}$