Surface Integrals Examples 3

# Surface Integrals Examples 3

Recall from the Surface Integrals page that the integral of a function $f$ over the surface $\delta$ defined parametrically by $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ is given by the following formula:

(1)
\begin{align} \quad \quad \iint_{\delta} f(x, y, z) \: dS = \iint_D f(\vec{r}(u, v)) \biggr \| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \biggr \| \: du \: dv = \iint_D f(x(u, v), y(u, v), z(u, v)) \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}

Also recall that if $\delta$ has a one-to-one projection onto the $xy$-plane, and $z = g(x, y)$ (which is equivalent to $G(x, y, z) = z - g(x, y) = 0$) generates our surface $\delta$ then the surface integral of $f$ over $\delta$ can be computed as:

(2)
\begin{align} \quad \iint_{\delta} f(x, y, z) \: dS = \iint_D f(x, y, g(x, y)) \frac{\| \nabla G(x, y, z) \|}{\biggr \| \frac{\partial G}{\partial z} \biggr \|} \: dx \: dy \end{align}

We will now look at some examples of evaluating surface integrals.

## Example 1

Evaluate the surface integral $\iint_{\delta}y \: dS$ where $\delta$ is the helicoid given parametrically by $\vec{r}(u, v) = (u \cos v, u \sin v, v)$ for $0 ≤ u ≤ 1$ and $0 ≤ v ≤ \pi$.

The surface $\delta$ of the helicoid is given below:

The surface area element $dS$ can be obtained nicely as:

(3)
\begin{align} \quad dS = \biggr \| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \biggr \| \: du \: dv \end{align}

Let's first compute $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$:

(4)
\begin{align} \quad \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \begin{vmatrix} \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{vmatrix} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \cos v & \sin v & 0\\ -u \sin v & u \cos v & 1 \end{vmatrix} = \sin v \vec{i} - \cos v \vec{j} + (u \cos^2 v + u \sin^2 v) \vec{k} = \sin v \vec{i} - \cos v \vec{j} + u \vec{k} \end{align}

Therefore we have that:

(5)
\begin{align} \quad dS = \sqrt{(\sin v)^2 + (-\cos v)^2 + (u)^2} \: du \: dv = \sqrt{\sin^2 v + \cos^2 v + u^2} \: du \: dv= \sqrt{1 + u^2} \: du \: dv \end{align}

We note that $D = \{ (u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ \pi \}$ and so we have that:

(6)
\begin{align} \quad \iint_{\delta} y \: dS = \iint_D u \sin v \sqrt{1 + u^2} \: du \: dv \\ \quad \iint_{\delta} y \: dS = \int_0^1 \int_0^{\pi} u \sin v \sqrt{1 + u^2} \: dv \: du \\ \quad \iint_{\delta} y \: dS = \int_0^1 u \sqrt{1 + u^2} \left [ - \cos v \right ]_{v=0}^{v=\pi} \: du \\ \quad \iint_{\delta} y \: dS = 2 \int_0^1 u \sqrt{1 + u^2} \: du \end{align}

Now we will need to use substitution to evaluate the integral on the righthand side of the equation. Let $m = 1 + u^2$. Then $dm = 2u \: du$ so $\frac{1}{2} dm = u \: du$ and so:

(7)
\begin{align} \quad \int u \sqrt{1 + u^2} \: du = \frac{1}{2} \int \sqrt{m} \: dm = \frac{1}{2} \frac{m^{3/2}}{3/2} = \frac{1}{3} m^{3/2} = \frac{1}{3} (1 + u^2)^{3/2} \end{align}

Thus:

(8)
\begin{align} \quad \iint_{\delta} y \: dS = \frac{2}{3} \left [ (1 + u^2)^{3/2} \right ]_{u=0}^{u=1} \\ \quad \iint_{\delta} y \: dS = \frac{2}{3} (2 \sqrt{2} - 1) \\ \quad \iint_{\delta} y \: dS = \frac{2(2 \sqrt{2} - 1)}{3} \end{align}

## Example 2

Evaluate the surface integral $\iint_{\delta} y \: dS$ where $\delta$ is the surface generated by the function $z = \frac{2}{3}(x^{3/2} + y^{3/2})$ for $0 ≤ x ≤ 1$ and $0 ≤ y ≤ 1$.

The surface $\delta$ is shown below:

Let $G(x, y, z) = z - \frac{2}{3}(x^{3/2} + y^{3/2})$. Our surface $\delta$ has a one-to-one projection onto the $xy$-plane, and so the area element $dS$ is given by:

(9)
\begin{align} \quad dS = \frac{\| \nabla G(x, y, z) \|}{\biggr \| \frac{\partial G}{\partial z} \biggr \|} \: dx \: dy \end{align}

We have that $\nabla G(x, y, z) = (-\sqrt{x}, -\sqrt{y}, 1)$ and $\frac{\partial G}{\partial z} = 1$. Therefore:

(10)
\begin{align} \quad dS = \frac{\sqrt{(-\sqrt{x})^2 + (-\sqrt{x})^2 + (1)^2}}{1} \: dx \: dy = \sqrt{x + x + 1} \: dx \: dy = \sqrt{2x + 1} \: dx \: dy \end{align}

We have that $D = \{ (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 \}$ and so therefore:

(11)
\begin{align} \quad \iint_{\delta} y \: dS = \iint_D y \sqrt{2x + 1} \: dx \: dy \\ \quad \iint_{\delta} y \: dS = \int_0^1 \int_0^1 y \sqrt{2x +1} \: dy \: dx \\ \quad \iint_{\delta} y \: dS = \int_0^1 \sqrt{2x + 1} \left [ \frac{y^2}{2} \right ]_{y=0}^{y=1} \: dx \\ \quad \iint_{\delta} y \: dS = \frac{1}{2} \int_0^1 \sqrt{2x + 1} \: dx \\ \quad \iint_{\delta} y \: dS = \frac{1}{2} \int_0^1 (2x + 1)^{1/2} \: dx \\ \quad \iint_{\delta} y \: dS = \frac{1}{2} \left [ \frac{1}{3} (2x + 1)^{3/2} \right ]_{x=0}^{x=1} \\ \quad \iint_{\delta} y \: dS = \frac{3\sqrt{3} - 1}{6} \end{align}