710595…: “Prior to the sour matrix”

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Consider the torus given by the following parametric formula: $\begin{align*} x(\theta,\phi) &= (3+\cos(\theta))\cos(\phi)\\ y(\theta,\phi) &= (3+\cos(\theta))\sin(\phi)\\ z(\theta,\phi) &= \sin(\theta) \end{align*}$ Let $\vec {T}(\theta ,\phi ) = \vector {x(\theta ,\phi ),y(\theta ,\phi ),z(\theta ,\phi )}$ Use a surface integral to compute the surface area of this torus. $\text {Area} = \answer {12\pi ^2}$

Start by computing: $\pp [\vec {T}]{\theta }\quad \text {and}\quad \pp [\vec {T}]{\phi }$

You should find that: $\pp [\vec {T}]{\theta } = \vector {-\sin (\theta )\cos (\phi ),-\sin (\theta )\sin (\phi ),\cos (\theta )}$

You should also find that: $\pp [\vec {T}]{\phi } = \vector {-(3+\cos (\theta ))\sin (\phi ),(3+\cos (\theta ))\cos (\phi ),0}$

Now compute: $\left |\pp [\vec {T}]{\theta }\cross \pp [\vec {T}]{\phi }\right |$

You should find that (after some simplification!): $\left |\pp [\vec {T}]{\theta }\cross \pp [\vec {T}]{\phi }\right |= 3+\cos (\theta )$

Finally, compute: $\int _0^{2\pi }\int _0^{2\pi } \left (3+ \cos (\theta ) \right )\d \phi \d \theta$