M344: Calculus III
Section 16.6
Surfaces, revisited¶
Recall that it is sometimes possible to parametrize a portion of a surface $S$ in $\mathbb{R}^3$ by a vector function $\mathbf{r}$ with two parameters.
$$
\mathbf{r}(u,v) = \big\langle x(u,v), y(u,v), z(u,v) \big\rangle.
$$
For examples, refer back to the notes from section 13.1.
Tangent Planes¶
Suppose $S$ is a surface in $\mathbb{R}^3$ parametrized by a vector function $\mathbf{r}(u,v)$, and suppose $P$ is a point on $S$ given by $\mathbf{r}(u_0,v_0)$. Then the tangent plane to $S$ at $P$, denoted by $T_PS$, is spanned by two vectors at $P$.
$$
\begin{array}{}
\mathbf{r}_u(u_0,v_0) & = & \dfrac{\partial \mathbf{r}}{\partial u}(u_0,v_0) & = & \left\langle \dfrac{\partial x}{\partial u} (u_0,v_0), \dfrac{\partial y}{\partial u} (u_0,v_0), \dfrac{\partial z}{\partial u} (u_0,v_0) \right\rangle, \quad \text{and} \\[2 ex]
\mathbf{r}_v(u_0,v_0) & = & \dfrac{\partial \mathbf{r}}{\partial v}(u_0,v_0) & = & \left\langle \dfrac{\partial x}{\partial v} (u_0,v_0), \dfrac{\partial y}{\partial v} (u_0,v_0), \dfrac{\partial z}{\partial v} (u_0,v_0) \right\rangle.
\end{array}
$$
The normal vector of the tangent plane at each point in $S$ is then given by
$$
\nu = \mathbf{r}_u \times \mathbf{r}_v.
$$
A surface $S$ is called smooth when $\Vert \nu\Vert \neq 0$ at every point in $S$. In this case, it is useful to have notation for the unit normal vector to a surface at each point. For smooth surfaces $S$ we define
$$
\mathbf{n} = \dfrac{\nu}{\Vert\nu\Vert}.
$$
The Gauss map of a smooth surface $S$ is the unit vector field defined by
$$
g_S(P) = \mathbf{n}(P),
$$
at all points $P$ in $S$. We won't use the Gauss map much in this class, but you will almost certainly see it in applications in your major fields.
Surface Area¶
Our main goal in this section is to reconsider the surface area of a surface in terms of a parametrization.
Suppose $S$ is a smooth surface given by a smooth parametrization $\mathbf{r}(u,v)$, and $S$ is covered just once as $(u,v)$ varies throughout the parameter domain $D$. We derive the following formula for the surface area of $S$.
$$
\text{Area}(S) = \int\!\!\!\!\int_S dS = \int\!\!\!\!\int_D \Vert \nu \Vert\, dA.
$$
Video lecture
1. Example¶
Give a parametrization of the sphere of radius $a > 0$ that satisfies the assumptions outlined in the discussion above, then find its surface area.
Video solution
2. Example¶
Find the surface area of the surface with parametric equations $x = u^2$, $y = uv$, and $z = \tfrac{1}{2}v^2$ on the rectangle $0\leq u \leq 1$, $0\leq v\leq 2$.
Video solution
3. Example¶
Compute the surface area of the torus obtained by rotating the circle in the $xz$-plane centered at $(R,0,0)$ with radius $r$, $0<r<R$, about the $z$-axis.
Hint: The first step is to think of a way to parametrize the torus.
Video solution
4. Example¶
Show that if $S$ is given by the graph of a function over a domain $D$, then the surface area of $S$ is given by the same formula we derived in chapter 15.
Video solution
5. Example¶
Compute the surface area of part of the plane $3x + 2y + z = 6$ that lies in the first octant.
Video solution
Discussion¶
Questions? You can ask them here.