M344: Calculus III
Section 16.7
Surface Integrals¶
Suppose $S$ is a smooth surface parametrized by a smooth vector fuction $\mathbf{r} = \mathbf{r}(u,v)$, and let $f$ be a function defined on a domain $E$ in $\mathbb{R}^3$ that contains the surface $S$. It make sense to ask if we can integrate the function $f$ along $S$.
Video lecture
The surface integral of $f$ along $S$ is given by
$$
\int\!\!\!\!\int_S f(x,y,z)\,dS = \int\!\!\!\!\int_D f\big(
\mathbf{r}(u,v)\big)\,\Vert\nu(u,v)\Vert\,dA,
$$
where $D$ is the parameter domain in the $uv$-plane for which $\mathbf{r}$ covers the surface $S$ exactly once as $(u,v)$ varies in $D$.
1. Example¶
Suppose that $f(x,y,z) =g\big(\sqrt{x^2 + y^2 + z^2\, }\big)$ where $g$ is a function of one variable such that $g(2) = -5$. Evaluate
$$
\int\!\!\!\!\int_S f(x,y,z)\, dS
$$
Where $S$ is the sphere $x^2 + y^2 + z^2 = 4$.
Video solution
2. Example¶
Evaluate
$$
\int\!\!\!\!\int_S y^2z^2\, dS,
$$
where $S$ is the portion of the cone $y = \sqrt{x^2 + z^2}$ given by $0 \leq y \leq 5$.
Video lecture
Orientable Surfaces¶
A smooth surface $S$ in $\mathbf{R}^3$ is said to be orientable if and only if it has a continuous Gauss map.
Video lecture, part I
Video lecture, part II
If $S$ is a smooth, orientable, closed surface in $\mathbb{R}^3$, then we may unambiguously refer to the surface's orientation as either positive or negative. We will say that such a surface is positively oriented if the normal vector always points in the "outward" direction; that is, away from the bounded region enclosed by the surface. The surface will be said to be negatively oriented if the normal vectors always point inward toward the bounded region.
Surface Integrals of Vector Fields¶
Suppse $S$ is an orientd surface with unit normal vector field $\mathbf{n}$, and suppose $\mathbf{F}(x,y,z)$ is a vector field defined on a region $E$ in $\mathbb{R}^3$ that contains the surface $S$.
The flux of $\mathbf{F}$ across $S$ is defined by the integral
$$
\int\!\!\!\!\int_S \mathbf{F}\cdot \mathbf{n}\,dS = \int\!\!\!\!\int_S \mathbf{F}\cdot d\mathbf{S}.
$$
If $S$ is parametrized by a smooth vector function $\mathbf{r}$ such that $S$ is traced out exactly once by $\mathbf{r}$ over a domain $D$, then the flux may be computed in the parameter domain as
$$
\int\!\!\!\!\int_S \mathbf{F}\cdot d\mathbf{S} = \int\!\!\!\!\int_D \mathbf{F}\cdot \nu\, dA.
$$
The details of this equivalence are left as an exercise for the interested reader.
Geometric Interpretation¶
If we regard $\mathbf{F}$ as the velocity field of a fluid flow, and the surface as a kind of cross-section in the flow that does not impede the flow in any way (like a thin net cast in a river), then the flux of the flow field represents the mass of fluid per unit of time moving past the surface, in the direction normal to the surface.
The Hairy Ball Theorem¶
There is an interesting theorem in Topology that says a continuous vector field defined on a sphere $S$ must vanish at at least one point. The creative topologists, geometers, and physicists that have come before us have imagined this theorem as saying that there is no way to comb a hairy ball without creating a cowlick.
1 minute video excursion
3. Example¶
Compute the flux of $\mathbf{F}$ across $S$, where
$$
\mathbf{F}(x,y,z) = xy\,\mathbf{i} + yz\,\mathbf{j} + xz\,\mathbf{k},
$$
and $S$ is the part of the paraboloid $z = 4 - x^2 - y^2$ that lies above the square $[0,1] \times [0,1]$ in the $xy$-plane.
Video solution
4. Example¶
The temperature at the point $(x,y,z)$ in a substance with conductivity $K = 6.5$ is
$$
u(x,y,z) = 2y^2 + 2z^2.
$$
Find the rate of heat flow inward across the cylindrical surface $y^2 + z^2 = 6$, $0 \leq x \leq 4$.
Video solution
Discussion¶
Questions? You can ask them here.