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Find the area of the surface given by the parametrization:
$\vec r(u,v)=((2+\cos(u))\cos(v),(2+\cos(u))\sin(v), \sin(u))$ while $0 \le u,v \le 2\pi$.

I have tried to solve this question using the given parametrization in the normal way by finding $\vec r_u \times \vec r_v$.
But things got really complicated with all the $\cos, \sin$ with two different parameters.
I was wondering if there's any cool tricks or other ways I couldn't see in order to solve this question.

Thanks in advance.
Note: Final answer is $8\pi^2$

Pwaol
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    It's a torus. Search in this site for "torus surface", it is plenty of useful results. At least five ways of make it easier. – Rafa Budría Jul 08 '21 at 15:11
  • @RafaBudría Thank you, I haven't tbh heard of that, I will try to read about it. but is there any way to make it easier without knowing it's a torus? since I received this question without ever hearing about a torus, or it was intended to be solved normally? – Pwaol Jul 08 '21 at 15:13
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    Following the, say, general method is not that messy. Things simplify a lot very quickly. Do the cross product group terms and use the trigonometric identity. Then calculate its norm, group and simplify. The final expression is really simple. – Rafa Budría Jul 08 '21 at 15:25
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    this parametrization is not bad to go about the surface integral. However you can check this - https://math.stackexchange.com/questions/1315349/surface-area-of-a-torus – Math Lover Jul 08 '21 at 15:46
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    Here is one that uses the same parametrization. You can see the steps and simplifications. https://math.stackexchange.com/questions/83269/surface-area-of-torus-of-revolution – Math Lover Jul 08 '21 at 15:49

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