Let $T$ be the torus obtained by revolving the circle
{$(x,0,z)| (x-3)^2 + z^2 = 1$} about the $z$-axis.
Find the area of the surface obtained by taking the intersection of $T$ with the cylinder {$(x,y,z)|x^2 + y^2 \leq 9$}.
i am not sure if the domain is right seen below in the trial
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my trial :
I parametrized the torus : $r(\theta,\phi) = (~(\cos(\theta)+3)\cos(\phi),(\cos(\theta)+3)\sin(\phi),\sin(\theta)~)$
the intersection is when :
${[(\cos(\theta)+3)\cos(\phi)]}^{2}+{[(\cos(\theta)+3)\sin(\phi)]}^2 \leq 9$
we get : $\frac{\pi}{2}\leq \theta \leq \frac{3\pi}{2}$
$0\leq \phi \leq 2\pi$
and we continue with the domain above :
|| $\frac{\partial r}{\partial \theta}$ x $\frac{\partial r}{\partial \phi}$ || = $(\cos(\theta)+3)$
and so :
S = $\iint ds$ = $\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (\cos(\theta)+3) {~d\theta}{~d\phi}$