# Integral on a surface

Let $$S$$ be a surface parametrized by the function $$\varphi (u,v)=(x(u,v),y(u,v),z(u,v))$$, and $$f$$ a function defined in all the points of the surface, then:

If $$f$$ is a scalar field (so, if $$f (x, y, z)$$ belongs to the real line)

Then,$$\displaystyle \int_S f\cdot dL=\int_D f(\varphi(u,v)) \cdot ||T_u \times T_v|| \ dudv$$$where, $$\displaystyle \begin{array}{l} T_u=\Big( \frac{d}{du} x(u,v), \frac{d}{du}y(u,v), \frac{d}{du}z(u,v) \\ T_v=\Big( \frac{d}{dv} x(u,v), \frac{d}{dv}y(u,v), \frac{d}{dv}z(u,v) \Big)\end{array}$$$

and $$D$$ is a region of the real plane where $$\varphi$$ is defined.

If $$F$$ is a vector field, $$F(x,y,z)=(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z))$$

Then, $$\displaystyle \int_S F \cdot dS=\int_Df(\varphi(u,v)) \cdot (T_u \times T_v) \ dudv$$\$ Namely, the integral of $$F$$ in the surface $$S$$ is the scalar product of the composed function with the parametrization times the vector product of $$T_u$$ and $$T_v$$.

## Procedure:

1. Take the parametrization of the surface $$S$$, and compute its vectors $$T_u$$, $$T_v$$. Therefore, compute the vector product.
2. Substitute $$x$$, $$y$$ and $$z$$ by $$x (u, v)$$, $$y(u, v)$$ and $$z (u, v)$$ in the function $$F$$, in accordance with the given parametrization.
3. Calculate the scalar product with the results of steps 1 and 2.
4. Calculate the final integral.