Lecture 52: Surface Integrals with Respect to Coordinates

60.1 Projection of Directed Surfaces and Surface Elements

Two-Sided Surface

One-Sided Surface

An oriented surface is a surface with a specified side; its orientation is represented by a chosen normal direction.

Let $\Sigma$ be an oriented surface. The signed projection of a surface element $\Delta S$ onto the $xOy$ plane is denoted by $(\Delta S)_{xy}$.

#####Definition: SurfaceIntegralOfTheSecondKind

description: $$\iint_{\Sigma}R(x,y,z)dxdy=\lim_{\lambda\to0}\sum_{i=1}^nR(\xi_i,\eta_i,\zeta_i)(\Delta S_i)_{xy}$$

Explanation

A surface integral of the second kind depends on the orientation of the surface.
Explanation:
The function value of a point $R(\xi_i,\eta_i,\zeta_i)$ multiplied by its projection on xy: $(\Delta S_i)_{xy}$
If it is $dxdy$ -> projection on xy;
If it is $dzdx$ -> projection on xz;
Reversing the orientation changes the sign:

\iint_{-\Sigma}(P\,dy\,dz+Q\,dz\,dx+R\,dx\,dy) =-\iint_{\Sigma}(P\,dy\,dz+Q\,dz\,dx+R\,dx\,dy).

#####Theorem: Calculationofthesecondtypeofcurvedareafraction

description:

Let the integrands be continuous on the smooth surface $\Sigma$.

1. If $\Sigma:z=z(x,y)$ over $D_{xy}$, then

$$\iint_{\Sigma}R\,dx\,dy=\pm\iint_{D_{xy}}R(x,y,z(x,y))\,dx\,dy.$$

Use $+$ for the upward orientation and $-$ for the downward orientation.

2. If $\Sigma:x=x(y,z)$ over $D_{yz}$, then

$$\iint_{\Sigma}P\,dy\,dz=\pm\iint_{D_{yz}}P(x(y,z),y,z)\,dy\,dz.$$

Use $+$ when the normal points toward positive $x$ and $-$ otherwise.

3. If $\Sigma:y=y(z,x)$ over $D_{zx}$, then

$$\iint_{\Sigma}Q\,dz\,dx=\pm\iint_{D_{zx}}Q(x,y(z,x),z)\,dz\,dx.$$

Use $+$ when the normal points toward positive $y$ and $-$ otherwise.

Explanation

Convert the surface integral to a double integral over the appropriate projection domain.
Selection of positive and negative signs:
If projecting onto xy:
Make the integral -> positive sign on the upper side;
Score the -> minus sign on the lower side;
If projecting onto yz:
The front side of yoz -> positive sign;
The -> minus sign behind yoz;
If projecting onto xz:
xoz -> to the right of the positive sign;
The left side of xoz -> minus;
Note:
If at z=f(x,y):
If x is a constant -> the projected field is a line, then the integral equals 0;
If y is a constant -> the projection field is also a line, then the integral equals 0;

#####Theorem: GaussDivergenceTheorem

Description: If the closed surface $\Sigma=\partial\Omega$ is oriented outward, then

$$\iint_{\Sigma}\left(P\,dy\,dz+Q\,dz\,dx+R\,dx\,dy\right)

=\iiint_{\Omega}\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dV.$$

Explanation

Concept:
Gauss's theorem relates the outward flux through a closed surface to the triple integral of the divergence over the enclosed volume.
It is the three-dimensional analogue of Green's theorem.

Supplement: Use the Gaussian formula to cover the surface

When the surface is not closed, add a face using the Gaussian formula;
Then, the result calculated using Gauss's formula - is added to the added part;

Concept: Conversion formula

Formula: $$\iint_{\Sigma}(P\cos\alpha+Q\cos\beta+R\cos\gamma)\mathrm{d}S=\iint_{\Sigma}(P\mathrm{d}y\mathrm{d}z+Q\mathrm{d}z\mathrm{d}x+R\mathrm{d}x\mathrm{d}y)$$
Explanation:
The left-hand side is a surface integral of the first kind.
The right-hand side is a surface integral of the second kind.
$\cos\alpha$, $\cos\beta$, and $\cos\gamma$ are the direction cosines of the chosen unit normal.