Surface Integrals over Basic Surfaces Examples 1

# Surface Integrals over Basic Surfaces Examples 1

Recall from the Surface Integrals over Basic Surfaces page that if $w = f(x, y, z)$ is a three variable bounded real-valued function over the smooth surface $\delta$ (with finite area) in $\mathbb{R}^3$ generated by $z = g(x, y)$ over $D$, then the surface integral of $f$ over $\delta$ is given by the formula:

(1)
\begin{align} \quad \iint_{\delta} f(x, y, z) \: dS = \iint_D f(x, y, g(x, y)) \sqrt{ \left ( \frac{\partial g}{\partial x} \right)^2 + \left ( \frac{\partial g}{\partial y} \right )^2 + 1} \: dA \end{align}

Let's look at some examples of evaluating surface integrals of surfaces $z = g(x, y)$

## Example 1

Evaluate the surface integral $\iint_{\delta} (xy + z) \: dS$ where $\delta$ is the portion of the plane $x + y + z = 2$ that lies in the first octant.

If $\delta$ is the portion of the plane $x + y + z = 2$ that lies in the first octant, then let $g(x, y) = 2 - x - y$ represent this plane. The partial derivatives of $g$ are:

(2)
\begin{align} \quad \frac{\partial g}{\partial x} = -1 \\ \quad \frac{\partial g}{\partial y} = -1 \end{align}

Therefore we have that:

(3)
\begin{align} \quad \iint_{\delta} (xy + z) \: dS = \iint_D (xy + (2 - x - y)) \sqrt{(-1)^2 + (-1)^2 + 1} \: dA \end{align}

Now we need to determine the region $D$ which lies on the $xy$-plane. If the plane, $g$, must lie in the first octant, then $x ≥ 0$ and $y ≥ 0$. We note that $g(x, y) = 0$ when $2 - x - y = 0$, that is, $g$ intersects the $xy$-plane on the line $y = -x + 2$, and so this line $y = -x + 2$ will be on the boundary of our region $D$:

We now see that the region $D$ which generates the surface $\delta$ from the function $g$ is given as $D = \{ (x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ -x + 2$.

Therefore we have that:

(4)
\begin{align} \quad \iint_{\delta} (xy + z) \: dS = \int_0^2 \int_0^{-x + 2} [xy + (2 - x - y)] \sqrt{(-1)^2 + (-1)^2 + 1} \: dy \: dx \\ \quad \iint_{\delta} (xy + z) \: dS = \int_0^2 \int_0^{-x + 2} \sqrt{3} ((x-1)y + 2 - x ) \: dy \: dx \\ \quad \iint_{\delta} (xy + z) \: dS = \int_0^2 \sqrt{3} \left [ \frac{(x - 1)y^2}{2} + (2-x)y \right ]_{y=0}^{y=-x + 2} \: dx \\ \quad \iint_{\delta} (xy + z) \: dS = \int_0^2 \sqrt{3} \frac{(x - 1)(2 - x)^2}{2} + (2 - x)^2 \: dx \\ \quad \iint_{\delta} (xy + z) \: dS = \sqrt{3} \int_0^2 \frac{1}{2} (4x -4x^2 +x^3 -4 +4x -x^2) + 4 - 4x + x^2 \: dx \\ \quad \iint_{\delta} (xy + z) \: dS = \sqrt{3} \int_0^2 2x - 2x^2 + \frac{x^3}{2} - 2 + 2x - \frac{x^2}{2} + 4 -4x + x^2 \: x \\ \quad \iint_{\delta} (xy + z) \: dS = \sqrt{3} \int_0^2 \frac{x^3}{2} -\frac{3x^2}{2} + 2 \: dx \\ \quad \iint_{\delta} (xy + z) \: dS = \sqrt{3} \left [\frac{x^4}{8} - \frac{x^3}{2} + 2x \right ]_0^2 \\ \quad \iint_{\delta} (xy + z) \: dS = \sqrt{3} \left ( 2 - 4 + 4 \right ) \\ \quad \iint_{\delta} (xy + z) \: dS = 2 \sqrt{3} \end{align}

## Example 2

Evaluate the surface integral $\iint_{\delta} z \: dS$ where $\delta$ is the upper half portion of the sphere centered at the origin with radius $2$.

If $\delta$ is the upper half portion of the sphere centered at the origin with radius $2$, then $g(x, y) = \sqrt{4 - x^2 - y^2} = (4 - x^2 - y^2)^{1/2}$ represents $\delta$. The partial derivatives of $g$ are:

(5)
\begin{align} \quad \frac{\partial g}{\partial x} = \frac{-2x}{2 \sqrt{4 - x^2 - y^2}} = \frac{-x}{\sqrt{4 - x^2 - y^2}} \\ \quad \frac{\partial g}{\partial y} = \frac{-2y}{2 \sqrt{4 - x^2 - y^2}} = \frac{-y}{\sqrt{4 - x^2 - y^2}} \end{align}

Therefore we have that:

(6)
\begin{align} \quad \iint_{\delta} z \: dS = \iint_D \sqrt{4 - x^2 - y^2} \sqrt{ \left ( \frac{-x}{\sqrt{4 - x^2 - y^2}} \right )^2 + \left ( \frac{-y}{\sqrt{4 - x^2 - y^2}} \right )^2 + 1} \: dA \\ \quad \iint_{\delta} z \: dS = \iint_D \sqrt{4 - x^2 - y^2} \sqrt{ \frac{x^2}{4 - x^2 - y^2} + \frac{y^2}{4 - x^2 - y^2} + 1} \: dA \end{align}

Now the region $D$ which generates the surface $\delta$ from $g$ can be described much more easily with polar coordinates as $D = \{ (r, \theta) : 0 ≤ r ≤ 2, 0 ≤ \theta ≤ 2 \pi \}$:

Therefore we have that:

(7)
\begin{align} \quad \iint_{\delta} z \: dS = \int_0^{2\pi} \int_0^2 \sqrt{4 - r^2} \left (\sqrt{ \frac{r^2 \cos^2 \theta}{4 - r^2} + \frac{r^2 \sin^2 \theta}{4 - r^2} + 1} \right ) r \: dr \: d \theta \\ \quad \iint_{\delta} z \: dS = \int_0^{2\pi} \int_0^2 \sqrt{4 - r^2} \left ( \sqrt{ \frac{r^2}{4 - r^2} + 1} \right ) r \: dr \: d \theta \\ \quad \iint_{\delta} z \: dS = \int_0^{2\pi} \int_0^2 \sqrt{4 - r^2} \left ( \frac{\sqrt{r^2 + 4 - r^2}}{\sqrt{4 - r^2}} \right ) r \: dr \: d \theta \\ \quad \iint_{\delta} z \: dS = \int_0^{2\pi} \int_0^2 \sqrt{4 - r^2} \frac{\sqrt{4}}{\sqrt{4 - r^2}} \: r \: dr \: d \theta \\ \quad \iint_{\delta} z \: dS = \int_0^{2\pi} \int_0^2 2r \: dr \: d \theta \\ \quad \iint_{\delta} z \: dS = \int_0^{2\pi} [r^2]_{r=0}^{r=2} \: d \theta \\ \quad \iint_{\delta} z \: dS = \int_0^{2\pi} 4 \: d \theta \\ \quad \iint_{\delta} z \: dS = [4 \theta]_{\theta=0}^{\theta= 2\pi} \\ \quad \iint_{\delta} z \: dS = 8 \pi \end{align}