Problems with surface area of the torus

[r1luher0]

Hello EvErybody...


A torus of radius 3 (and cross-sectional radius 1) can be represented parametrically by the function ?: D ? \mathbbR3

by: ?(?, ?) = ((3 + cos?)cos?,(3 +cos?)sin?, sin?)
where D is the rectangle given by 0 ? ? ? 2?, 0 ? ? ? 2?.

help Please.

 

[galactus]

I will going to use u and v instead of theta and phi

Begin by calculating \(\displaystyle r_{u}, \;\ r_{v}\)

\(\displaystyle r_{u}=-sin(u)cos(v)i-sin(u)sin(v)j+cos(u)k\)

\(\displaystyle r_{v}=-(3+cos(u))sin(v)i+(3+cos(u))cos(v)j\)

Now, find the cross product:

\(\displaystyle r_{u}\times r_{v} = \begin{vmatrix}-sin(u)cos(v)&-sin(u)sin(v)&+cos(u)\\-(3+cos(u))sin(v)&(3+cos(u))cos(v)&0\end{vmatrix}\)

\(\displaystyle -(3+cos(u))(cos(v)cos(u)i+sin(v)cos(u)j+sin(u)k)\)

This implies:

\(\displaystyle ||r_{u}\times r_{v}||=(3+cos(u))\sqrt{(cos(v)cos(u))^{2}+(sin(v)cos(u))^{2}+sin^{2}(u)}=3+cos(u)\)

The surface area is \(\displaystyle \int_{0}^{2\pi}\int_{0}^{2\pi}(3+cos(u))dudv\)

Solve the integral and you have your surface area of the torus.

 

[r1luher0]

Thankssss for Your Help My FrIend...


You SaVEd my liFe..

Muchas Gracias DEsde MExico!