Surface Integrals Review
Surface Integrals Review
We will now review some of the recent content regarding surface integrals.
- On the Parametric Surfaces we began to look at parametric surfaces. We saw that for $u$ and $v$ are parameters that we can define some neat surfaces as $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $a ≤ u ≤ b$ and $c ≤ v ≤ d$.
- We said that if two surfaces $S_1$ and $S_2$ join at a common boundary curve then the resulting surface is a Composite Surface. One example of a composite surface is a cube of six surface squares.
- We then looked at some techniques to parameterize surfaces on the Parameterizing Surfaces page.
- We noted that if a surface is given as a function $z = f(x, y)$, then we can easily parameterize this surface as $\vec{r}(x, y) = (x, y, f(x, y))$.
- We also saw that we could parameterize sphere and ellipsoids. For a general ellipse $ax^2 + by^2 + cz^2 = d^2$ where $a, b, c, d > 0$ we have the following parameterization in spherical coordinates:
\begin{align} \vec{r}(\phi, \theta) = \left\{\begin{matrix} x(\phi, \theta) = \sqrt{\frac{d}{a}}\sin \phi \cos \theta\\ y(\phi, \theta) = \sqrt{\frac{d}{b}} \sin \phi \sin \theta \\ z(\phi, \theta) = \sqrt{\frac{d}{c}} \cos \phi \end{matrix}\right. \quad 0 ≤ \phi ≤ \pi \: , \: 0 ≤ \theta ≤ 2\pi \end{align}
- If a surface is generated by rotating a function $y = f(x)$ about the $x$-axis then we could parameterize the resulting surface as:
\begin{align} \quad \vec{r}(x, \theta) = \left\{\begin{matrix} x(x, \theta) = x \\ y(x, \theta) = f(x) \cos \theta\\ z(x, \theta) = f(x) \sin \theta \end{matrix}\right. \quad a ≤ x ≤ b \: , \: 0 ≤ \theta ≤ \alpha \end{align}
- Recall from the Surface Integrals page that if $\delta$ is a smooth surface defined parametrically by $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ and $w = f(x, y, z)$ is a three variable real-valued function, then the surface integral of $f$ over $\delta$ is given by the following formula:
\begin{align} \quad \quad \iint_{\delta} f(x, y, z) \: dS = \iint_D f(\vec{r}(u, v)) \biggr \| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \biggr \| \: du \: dv = \iint_D f(x(u, v), y(u, v), z(u, v)) \sqrt{ \left ( \frac{\partial (y, z)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (z, x)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (x, y)}{\partial (u, v)} \right )^2 } \: du \: dv \end{align}
- On the Surface Integrals over Basic Surfaces we saw that if the surface $\delta$ is given as a function $z = g(x, y)$ and $w = f(x, y, z)$ is a three variable real-valued function, then the surface integral of $f$ over the surface generated by $g$ is given by the formula:
\begin{align} \quad \iint_{\delta} f(x, y, z) \: dS = \iint_D f(x, y, g(x, y)) \sqrt{ \left ( \frac{\partial g}{\partial x} \right)^2 + \left ( \frac{\partial g}{\partial y} \right )^2 + 1} \: dx \: dy \end{align}
- Similar formulas can be derived when a surface $\delta$ is generated by a function $x = h(y, z)$ or $y = h(x, z)$.
- On the Computing Surface Area with Surface Integrals we also noted that if $f(x, y, z) = 1$ then the surface integral of $f$ over the surface $\delta$ will be equal to the area of $\delta$, that is:
\begin{align} \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \sqrt{ \left ( \frac{\partial (y, z)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (z, x)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (x, y)}{\partial (u, v)} \right )^2 } \: du \: dv \end{align}
- We then looked at Orientable Surfaces. We said that a surface was Orientable if there exists a unit normal field $\hat{N}(P)$ that is normal to every point $P$ on the surface and continuously varies over the surface.
- Since $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$ is normal to the surface $\delta$, then we can define a natural unit normal field for the surface $\delta$ as:
\begin{align} \quad \hat{N} = \frac{\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}}{\biggr \| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \biggr \|} \end{align}
- If a surface $\delta$ is generated by a function $z = g(x, y)$ then the natural unit normal field for $\delta$ is:
\begin{align} \quad \hat{N} = \frac{-\frac{\partial f}{\partial x} \vec{i} - \frac{\partial f}{\partial y} \vec{j} + \vec{k}}{\sqrt{\left ( \frac{\partial f}{\partial x} \right )^2 + \left ( \frac{\partial f}{\partial y} \right )^2 + 1}} \end{align}
- We then defined Surface Integrals of Vector Fields, also known as Flux Integrals. If $\delta$ is an orientable surface with unit normal field $\hat{N}$ and if $\mathbf{F}$ is a continuous vector field then the surface integral of $\mathbb{F}$ across $\delta$ (also known as the flux of $\mathbb{F}$ across $\delta$) is given by:
\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = \pm \iint_D \left (P(x, y, z) \frac{\partial (y, z)}{\partial (u, v)} + Q(x, y, z) \frac{\partial (z, x)}{\partial (u, v)} + R(x, y, z) \frac{\partial (x, y)}{\partial (u, v)} \right ) \: du \: dv \end{align}
- The plus/minus will vary depending on the orientation we want. An upward orientation requires that the last component of the unit normal field $\hat{N}$ is positive and a downward orientation requires that the last component of the unit normal field $\hat{N}$ is negative. We then choose the sign accordingly.