Surface Integrals Examples 4

# Surface Integrals Examples 4

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 $\iint_{\delta} (z + x^2 y) \: dS$ where $\delta$ is the portion of the cylinder $y^2 + z^2 = 1$ in the first octant lying between the planes $x = 0$ and $x = 3$.

The surface $\delta$ is depicted below:

This surface $\delta$ is a quarter cylinder of radius $1$ that is parallel to the $x$-axis. This cylinder can be nicely parameterized for $0 ≤ u ≤ 3$ and $0 ≤ v≤ \frac{\pi}{2}$:

(3)
\begin{align} \quad \vec{r}(u, v) = u \vec{i} + \cos v \vec{j} + \sin v \vec{k} \end{align}

The area element $dS$ is given by:

(4)
\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 compute $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$:

(5)
\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}\\ 1& 0 & 0\\ 0 & -\sin v& \cos v\\ \end{vmatrix} = 0 \vec{i} - \cos v \vec{j} + \sin v \vec{k} \end{align}

Therefore we have that:

(6)
\begin{align} \quad dS = \sqrt{(0)^2 + (-\cos v)^2 + (\sin v)^2} \: du \: dv = \sqrt{ \cos^2 v + \sin^2 v} \: du \: dv = 1 \: du \: dv \end{align}

For $D = \left \{ (u, v) : 0 ≤ u ≤ 3, 0 ≤ v ≤ \frac{\pi}{2} \right \}$ we have that:

(7)
\begin{align} \quad \iint_{\delta} (z + x^2 y) \: dS = \iint_D (\sin v + u^2 \cos v) \: du \: dv \\ \quad \iint_{\delta} (z + x^2 y) \: dS = \int_0^3 \int_0^{\pi/2} (\sin v + u^2 \cos v) \: dv \: du \\ \quad \iint_{\delta} (z + x^2 y) \: dS = \int_0^3 \left [ -\cos v + u^2 \sin v \right ]_{v=0}^{v=\pi/2} \: du \\ \quad \iint_{\delta} (z + x^2 y) \: dS = \int_0^3 (u^2 + 1) \: du \\ \quad \iint_{\delta} (z + x^2 y) \: dS = \left [ \frac{u^3}{3} + u \right ]_{u=0}^{u=3} \\ \quad \iint_{\delta} (z + x^2 y) \: dS = 12 \end{align}

## Example 2

Evaluate $\iint_{\delta} x \: dS$ where $\delta$ is the triangle with vertices $(1, 0, 0)$, $(0, -2, 0)$, and $(0, 0, 4)$.

This triangle $\delta$ must lie on a plane in $\mathbb{R}^3$. Recall that an equation of a plane can be obtained with a vector that is perpendicular to the plane, call it $\vec{n}$, and a point on the plane. We already have three points on this plane, so all we need is a vector that is perpendicular to this plane.

Consider the vectors $(0, -2, 0) - (1, 0, 0) = (-1, -2, 0)$ and $(0, 0, 4) - (1, 0, 0) = (-1, 0, 4)$. These vectors both lie on the plane containing the triangle $\delta$. Their cross product is perpendicular to the plane:

(8)
\begin{align} \quad (-1, 2, 0) \times (-1, 0, 4) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -1 & -2 & 0\\ -1 & 0 & 4 \end{vmatrix} = -8 \vec{i} + 4 \vec{j} - 2 \vec{k} \end{align}

Therefore the vector $(-4, 2, -1)$ is perpendicular to the plane containing the triangle $\delta$. The equation of this plane is:

(9)
\begin{align} \quad -4(x - 1) + 2y - z = 0 \\ \quad z = 4 - 4x + 2y \end{align}

This plane has a one-to-one projection onto the $xy$-plane. Let $G(x, y, z) = z - 4 + 4x - 2y = 0$. Then $\nabla G(x, y, z) = (4, -2, 1)$ and $\frac{\partial G}{\partial z} = 1$. Therefore the area element $dS$ is given by:

(10)
\begin{align} \quad dS = \sqrt{(4)^2 + (-2)^2 + (1)^2} \: dx \: dy = \sqrt{21} \: dx \: dy \end{align}

We now need to determine the boundaries for our surface.

Notice that the points $(1, 0, 0)$ and $(0, -2, 0)$ intersect the $xy$-plane since their $z$-coordinates are equal to zero. In the $xy$-plane, these points correspond to $(1, 0)$ and $(0, -2)$. The slope of the line connecting these two points is $\frac{2}{-1} = 2$. Plugging this into the equation $y = mx + b$ and using the point $(x, y) = (1, 0)$ we have that $0 = 2(1) + b$ which implies that $b = -2$, Thus the equation of this line is $y = 2x - 2$.

Now look at the points $(1, 0, 0)$ and $(0, 0, 4)$. When we project these points onto the $xy$ plane we get $(1, 0)$ and $(0, 0)$. The equation of the line that passes through these points is $y = 0$.

Lastly, we look at the points $(0, -2, 0)$ and $(0, 0, 4)$. When we project these points onto the $xy$ plane we get $(0, -2)$ and $(0, 0)$. The equation of the line that passes through these points is $x = 0$.

When we connect these lines, we get a projection of the triangle $\delta$ onto the $xy$-plane as graphed below:

This is our region $D$ and can be described as:

(11)
\begin{align} \quad D = \{ (x, y) : 0 ≤ x ≤ 1 : 2x - 2 ≤ y ≤ 0 \} \end{align}

Thus we have that:

(12)
\begin{align} \quad \iint_{\delta} x \: dS = \sqrt{21} \iint_D x \: dx \: dy \\ \quad \iint_{\delta} x \: dS = \sqrt{21} \int_0^1 \int_{2x - 2}^{0} x \: dy \: dx \\ \quad \iint_{\delta} x \: dS = \sqrt{21} \int_0^1 x \left [ y \right ]_{y=2x - 2}^{y=0} \: dx \\ \quad \iint_{\delta} x \: dS = \sqrt{21} \int_0^1 x (2 - 2x) \: dx \\ \quad \iint_{\delta} x \: dS = \sqrt{21} \int_0^1 (2x - 2x^2) \: dx \\ \quad \iint_{\delta} x \: dS = \sqrt{21} \left [ x^2 - \frac{2x^3}{3} \right ]_{x=0}^{x=1} \\ \quad \iint_{\delta} x \: dS = \sqrt{21} \left (1 - \frac{2}{3} \right ) \\ \quad \iint_{\delta} x \: dS = \frac{\sqrt{21}}{3} \end{align}