Part 19 - Surface integrals

In this part we’ll cover surface integrals.

Let’s first start with tangent planes to surfaces.

Suppose $S$ is a parametric surface with parameterization $r: D \subset \mathbb{R}^2 \rarr \mathbb{R}^3$ .

\vec{r}(u, v) = x(u, v) \vec{i} + y(u, v) \vec{j} + z(u, v) \vec{k}.

We define $\vec{r_u}(u, v)$ as:

\vec{r_u} = x_u(u, v) \vec{i} + y_u(u, v) \vec{j} + z_u(u, v) \vec{k}.

And for the partial derivative of $v$ :

\vec{r_v} = x_v(u, v) \vec{i} + y_v(u, v) \vec{j} + z_v(u, v) \vec{k}.

Definition 1 (Tangent plane)

The tangent plane, $\pi$ , to $S$ , at the point $r(u_0, v_0)$ is the plane that contains $r_u(u_0, v_0)$ and $r_v(u_0, v_0)$ .

Equivalently, it is the plane perpendicular to the cross product of $r_u(u_0, v_0)$ and $r_v(u_0, v_0)$ , which goes through $r(u_0, v_0)$ .

Example 1

Find the tangent plane to $x = u^2, y = v^2, z = u + 2v$ . At the point $(1, 1, 3)$ .

\begin{align*} \vec{r}(u, v) & = \langle u^2, v^2, u + 2v \rangle, \newline \vec{r_u}(u, v) & = \langle 2u, 0, 1 \rangle, \newline \vec{r_v}(u, v) & = \langle 0, 2v, 2 \rangle. \end{align*}

\begin{align*} \vec{r_u}(u, v) \times \vec{r_v}(u, v) & = \begin{vmatrix} i & j & k \newline 2u & 0 & 1 \newline 0 & 2v & 2 \end{vmatrix} \newline & = \langle (0 \cdot 2) - (1 \cdot 2v), -((2u \cdot 2) - (1 \cdot 0)), (2u \cdot 2v) - (0 \cdot 0) \rangle \newline & = \langle -2v, -4u, 4uv \rangle, \newline \vec{r_u}(1, 1) \times \vec{r_v}(1, 1) & = \langle -2, -4, 4 \rangle. \end{align*}

Therefore:

\begin{align*} \pi: -2(x - 1) - 4(y - 1) + 4(z - 3) & = 0, \newline -2x + 2 - 4y + 4 + 4z - 12 & = 0, \newline -x - 2y + 2z & = 3. \end{align*}

Let $r: D \rarr \mathbb{R}^3$ be a parameterization for a surface, $S$ .

Definition 2 (Surface area)

\text{Area}(S) = \iint_D | \vec{r_u} \times \vec{r_v} | \, du \, dv.

Example 2

Find the area of the paraboloid

z = x^2 + y^2

Below the plane $z = 9$ .

We can parameterize such as:

\begin{align*} \vec{r}(u, v) & = \langle u, v, u^2 + v^2 \rangle, \newline D & = \{(u, v) \in \mathbb{R}^2 \mid u^2 + v^2 \leq 9 \}. \end{align*}

\begin{align*} \vec{r_u}(u, v) & = \langle 1, 0, 2u \rangle, \newline \vec{r_v}(u, v) & = \langle 0, 1, 2v \rangle, \newline \vec{r_u} \times \vec{r_v} & = \begin{vmatrix} i & j & k \newline 1 & 0 & 2u \newline 0 & 1 & 2v \end{vmatrix} \newline & = \langle (0 \cdot 2v) - (2u \cdot 1), -((1 \cdot 2v) - (2u \cdot 0)), (1 \cdot 1) - (0 \cdot 0) \rangle \newline & = \langle -2u, -2v, 1 \rangle, \newline | \vec{r_u} \times \vec{r_v} | & = \sqrt{4u^2 + 4v^2 + 1}. \end{align*}

Let’s also change to polar coordinates, since we have a circle:

S = \{(r, \theta) \mid 0 \leq r \leq 3, 0 \leq \theta \leq 2\pi \}

\begin{align*} \iint_D | \vec{r_u} \times \vec{r_v} | \, du \, dv & = \iint_S \sqrt{4r^2 + 1} \cdot r \, dr \, d\theta \newline & = \int_0^{2\pi} \int_0^3 r\sqrt{4r^2 + 1} \, dr \, d\theta \newline & = 2\pi \int_0^3 r\sqrt{4r^2 + 1} \, dr \end{align*}

Let $u = 4r^2 + 1$ , $du = 8r \, dr$ , $u(0) = 1$ and $u(3) = 37$ .

\begin{align*} 2\pi \int_0^3 r\sqrt{4r^2 + 1} \, dr & = \dfrac{2\pi}{8} \int_1^{37} \sqrt{u} \, du \newline & = \dfrac{\pi}{4} \left[ \dfrac{2 u^{\frac{3}{2}}}{3} \right]_{u = 1}^{u = 37} \newline & = \dfrac{\pi}{4} \left( \dfrac{2 \cdot 37^{\frac{3}{2}}}{3} - \dfrac{2}{3} \right) \end{align*}

Yeah I’m just going to leave it at that…

Definition 3 (Surface integral)

Let $r: D \rarr \mathbb{R}^3$ be a parameterization of a surface, $S$ .

Let $f: S \rarr \mathbb{R}$ be a function, then:

\iint_S f(x, y, z) \, dS = \iint_D f(\vec{r}(u, v)) | \vec{r_u} \times \vec{r_v} | \, du \, dv.

Example 3

Compute $\iint_S x^2 \, dS$ where $S$ is the unit sphere in $\mathbb{R}^3$ .

S = \{(x, y, z) \mid x^2 + y^2 + z^2 = 1 \}

We could use spherical, but since we know that the radius is 1, $r$ is useless, therefore we can use “polar” coordinates.

\vec{r}(\phi, \theta) = \langle \sin(\phi) \cos(\theta), \sin(\phi) \sin(\theta), \cos(\phi) \rangle, \quad 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi.

Which means:

\begin{align*} \vec{r_{\phi}} & = \langle \cos(\phi) \cos(\theta), \cos(\phi) \sin(\theta), -\sin(\phi) \rangle, \newline \vec{r_{\theta}} & = \langle -\sin(\phi) \sin(\theta), \sin(\phi) \cos(\theta), 0 \rangle. \end{align*}

\begin{align*} \vec{r_{\phi}} \times \vec{r_{\theta}} & = \begin{vmatrix} i & j & k \newline \cos(\phi) \cos(\theta) & \cos(\phi) \sin(\theta) & -\sin(\phi) \newline -\sin(\phi) \sin(\theta) & \sin(\phi) \cos(\theta) & 0 \end{vmatrix} \newline & = \langle \sin^2(\phi) \cos(\theta), \sin^2(\phi) \sin(\theta), \cos^2(\theta) \cos(\phi) \sin(\phi) + \sin^2(\theta) \cos(\phi) \sin(\phi) \rangle, \newline | \vec{r_{\phi}} \times \vec{r_{\theta}} | & = \ldots = \sin(\phi). \end{align*}

CBA

Example 4

Evaluate $\iint_S y \, dS$ where $S = z = x + y^2$ for $0 \leq x \leq 1$ and $0 \leq y \leq 2$ .

\begin{align*} \vec{r}(u, v) & = \langle u, v, u + v^2 \rangle, \newline \vec{r_u}(u, v) & = \langle 1, 0, 1 \rangle, \newline \vec{r_v}(u, v) & = \langle 0, 1, 2v \rangle, \newline \vec{r_u} \times \vec{r_v} & = \begin{vmatrix} i & j & k \newline 1 & 0 & 1 \newline 0 & 1 & 2v \end{vmatrix} = \langle -1, -2v, 1 \rangle, \newline | \vec{r_u} \times \vec{r_v} | & = \sqrt{4v^2 + 2}. \end{align*}

\begin{align*} \iint_S y \, dS & = \int_0^1 \int_0^2 v \sqrt{4v^2 + 2} \, dv \, du \newline & = \int_0^2 v \sqrt{4v^2 + 2} \, dv \end{align*}

Let $u = 4v^2 + 2$ , $du = 8v \, dv$ , $u(0) = 2$ and $u(2) = 18$ .

\begin{align*} \dfrac{1}{8} \int_2^{18} \sqrt{u} \, du & = \dfrac{1}{8} \left[ \dfrac{2 u^{\frac{3}{2}}}{3} \right]_{u = 2}^{u = 18} \newline & = \dfrac{1}{8} \left[ \left(\dfrac{2 \cdot 18^{\frac{3}{2}}}{3} \right) - \left(\dfrac{2 \cdot 2^{\frac{3}{2}}}{3} \right) \right] \end{align*}

Good enough :]