Part 19 - Surface integrals

In this part we’ll cover surface integrals.

Let’s first start with tanget 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 derivate 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 $(1, 1, 3)$ .

\vec{r}(u, v) = \langle u^2, v^2, u + 2v \rangle

\vec{r_u}(u, v) = \langle 2u, 0, 1 \rangle

\vec{r_v}(u, v) = \langle 0, 2v, 2 \rangle

\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} = \langle (0 \cdot 2) - (1 \cdot 2v), -(2u \cdot 2) - (1 \cdot 0), (2u \cdot 2v) - (0 \cdot 0) \rangle = \langle -2v, -2u, 4uv \rangle

\vec{r_u}(1, 1) \times \vec{r_v}(1, 1) = \langle -2, -2, 4 \rangle

Therefore:

\pi = -2(x - 1) - 4(y - 1) + 4(z - 3)

-2x +2 - 4y + 4 + 4z - 12 = 0

-x -2y + 2z = 3

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} |\ dx\ dy

Example 2

Find the area of the paraboloid

z = x^2 + y^2

Below the plane $z = 9$ .

We can parameterize such as:

\vec{r}(u, v) = \langle u, v, u^2 + v^2 \rangle

D = \{(u, v) \in \mathbb{R}^2 | u^2 + v^2 \leq 9 \}

\vec{r_u}(u, v) = \langle 1, 0, 2u \rangle

\vec{r_v}(u, v) = \langle 0, 1, 2v \rangle

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

| \vec{r_u} \times \vec{r_v} | = \sqrt{4u^2 + 4v^2 + 1}

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

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

\iint_D | \vec{r_u} \times \vec{r_v} |\ du\ dv = \iint_S \sqrt{4r + 1} \cdot r\ dr\ d\theta = \int_0^{2\pi} \int_0^3 r\sqrt{4r + 1}\ dr\ d\theta = 2\pi \int_0^3 r\sqrt{4r + 1}\ dr

Let $u = 4r + 1$ , $du = 4\ dr$ , $u(0) = 1$ and $u(3) = 13$ .

2\pi \int_1^{13} r \sqrt{u}\ dr

2\pi \int_1^{13} (u - 1) \sqrt{u}\ du

2\pi \int_1^{13} u^{\frac{3}{2}} - \sqrt{u}\ du

2\pi \left[ \dfrac{2 u^{\frac{5}{2}}}{5} - \dfrac{2 u^{\frac{3}{2}}}{3} \bigg\rvert_{u = 1}^{u = 13} \right]

2\pi \left[ \dfrac{2 u^{\frac{5}{2}}}{5} - \dfrac{2 u^{\frac{3}{2}}}{3} \bigg\rvert_{u = 1}^{u = 13} \right]

2\pi \left[ \left(\dfrac{2 \cdot 13^{\frac{5}{2}}}{5} - \dfrac{2 \cdot 13^{\frac{3}{2}}}{3}\right) - \left(\dfrac{2 \cdot 1^{\frac{5}{2}}}{5} - \dfrac{2 \cdot 1^{\frac{3}{2}}}{3}\right) \right]

2\pi \left[ \left(\dfrac{2 \cdot 13^{\frac{5}{2}}}{5} - \dfrac{2 \cdot 13^{\frac{3}{2}}}{3}\right) - \left(\dfrac{2}{5} - \dfrac{2}{3}\right) \right]

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) | 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 \ | \ 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi

Which means:

\vec{r_{\phi}} = \langle cos(\phi)\ cos(\theta), cos(\phi)\ sin(\theta), -sin(\phi) \rangle

\vec{r_{\theta}} = \langle -sin(\phi)\ sin(\theta), sin(\phi)\ cos(\theta), 0 \rangle

\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}

\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

| \vec{r_{phi}} \times \vec{r_{\theta}} | = \ldots = sin(\phi)

CBA

Example 4

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

\vec{r}(u, v) = \langle u, v, u + v^2 \rangle

\vec{r_u}(u, v) = \langle 1, 0, 1 \rangle

\vec{r_v}(u, v) = \langle 0, 1, 2v \rangle

\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

| \vec{r_u} \times \vec{r_v} | = \sqrt{4v^2 + 2}

\iint_S y\ dS = \int_0^1 \int_0^2\ v \sqrt{4v^2 + 2}\ dv\ du

\int_0^2\ v \sqrt{4v^2 + 2}\ dv

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

\dfrac{1}{8} \int_2^{18}\ \sqrt{u}\ du

\dfrac{1}{8} \left[ \dfrac{2 u^{\frac{3}{2}}}{3} \bigg\rvert_{u = 2}^{u = 18} \right]

\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]

Good enough :]