Computing Surface Area with Surface Integrals Examples 2

# Computing Surface Area with Surface Integrals Examples 2

Recall from the Computing Surface Area with Surface Integrals page that if a surface $\delta$ is given parametrically as $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ then the surface area of $\delta$ is given by:

(1)
\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 will now look at some more examples of computing the surface area of a surface.

## Example 1

Find the surface area of the portion of the plane $x + 2y + 3z = 1$ inside the cylinder $x^2 + y^2 = 3$.

Note that we are parameterizing the surface $x + 2y + 3z = 1$ over the disk $D$, $x^2 + y^2 ≤ 3$. We can parameterize this surface nicely by letting $x = x$, $y = y$, and $z = \frac{1 - x - 2y}{3}$, and so $\vec{r}(x, y) = \left (x, y, \frac{1 - x - 2y}{3} \right )$ is a parameterization of our surface.

Let's compute $\frac{\partial \vec{r}}{\partial x} \times \frac{\partial \vec{r}}{\partial y}$.

(2)
\begin{align} \quad \frac{\partial \vec{r}}{\partial x} \times \frac{\partial \vec{r}}{\partial y} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial x}{\partial x} & \frac{\partial y}{\partial x} & \frac{\partial z}{\partial x} \\ \frac{\partial x}{\partial y} & \frac{\partial y}{\partial y} & \frac{\partial z}{\partial y} \end{bmatrix} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -\frac{1}{3} \\ 0 & 1 & -\frac{2}{3} \end{bmatrix} \end{align}

We immediately see that $\frac{\partial (y, z)}{\partial (x, y)} = \frac{1}{3}$, $\frac{\partial (z, x)}{\partial (x, y)} = \frac{2}{3}$ and $\frac{\partial (x, y)}{\partial (x, y)} = 1$, and thus:

(3)
\begin{align} \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \sqrt{ \left ( \frac{\partial (y, z)}{\partial (x, y)} \right )^2 + \left ( \frac{\partial (z, x)}{\partial (x, y)} \right )^2 + \left ( \frac{\partial (x, y)}{\partial (x, y)} \right )^2 } \: dx \: dy \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \sqrt{ \left ( \frac{1}{3} \right )^2 + \left ( \frac{2}{3} \right )^2 + (1)^2} \: dx \: dy \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \sqrt{\frac{14}{9}} \: dx \: dy \\ \end{align}

Using polar coordinates to represent $D = \{ (x, y) : x^2 + y^2 ≤ 3 \} = \{ (r, \theta) : 0 ≤ r ≤ \sqrt{3}, 0 ≤ \theta ≤ 2\pi \}$ and we have that:

(4)
\begin{align} \quad \mathrm{Surface \: Area \: of \: \delta} = \frac{\sqrt{14}}{3} \int_0^{2\pi} \int_0^{\sqrt{3}} r \: dr \: d \theta \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \frac{\sqrt{14}}{3} \int_0^{2\pi} \left [ \frac{r^2}{2} \right ]_{r = 0}^{r = \sqrt{3}} \: d \theta \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \frac{\sqrt{14}}{3} \int_0^{2\pi} \frac{3}{2} \: d \theta \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \frac{\sqrt{14}}{2} \left [ \theta \right ]_{\theta = 0}^{\theta = 2\pi} \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \sqrt{14} \pi \end{align}

## Example 2

Find the surface area of the surface given parameterically by $\vec{r}(u, v) = (u \cos v, u \sin v, v)$ for $0 ≤ u ≤ 1$ and $0 ≤ v ≤ \pi$.

We will first compute $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$ as follows:

(5)
\begin{align} \quad \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{bmatrix} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ \cos v & \sin v & 0 \\ -u \sin v & u \cos v & 1 \end{bmatrix} \end{align}

We immediately have that $\frac{\partial (y, z)}{\partial (u, v)} = \sin v$, $\frac{\partial (z, x)}{\partial (u, v)} = \cos v$ and $\frac{\partial (x, y)}{\partial (u, v)} = u \cos^2 v + u \sin^2 v = u$ and so:

(6)
\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 \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \sqrt{(\sin v)^2 + (\cos v)^2 + u^2} \: du \: dv \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \sqrt{1 + u^2} \: du \: dv \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \int_0^1 \int_0^{\pi} \sqrt{1 + u^2} \: dv \: du \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \int_0^1 \sqrt{1 + u^2} \left [ v \right ]_{v = 0}^{v = \pi} \: du \\ \quad \mathrm{Surface \: Area \: of \: \delta} = 2\pi \int_0^1 \sqrt{1 + u^2} \: du \\ \quad \mathrm{Surface \: Area \: of \: \delta} = 2\pi \left [ \frac{\ln \mid \sqrt{1 + u^2} + u \mid +u \sqrt{u^2 + 1}}{2} \right ]_{u =0}^{u=1} \: du \\ \quad \mathrm{Surface \: Area \: of \: \delta} = \pi \left (\ln (\sqrt{2} + 1) + \sqrt{2} \right ) \end{align}