Surface Integrals over Basic Surfaces

Surface Integrals over Basic Surfaces

Recall from the Surface Integrals page that if $w = f(x, y, z)$ is a three variable bounded real-valued function over the smooth finite area surface $\delta$ in $\mathbb{R}^3$ then the surface integral of $f$ over $\delta$ where $\delta$ is defined parametrically by $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ is given by:

\begin{align} \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}

The most basic type of surface integral that we can evaluate are surfaces $\delta$ defined as a two variable real-valued function $z = g(x, y)$. We can parameterize the function $z = g(x, y)$ easily by letting $x = x$, $y = y$, and $z = g(x, y)$ (which we saw on the Parametric Surfaces page).

Now suppose that the first partial derivatives of $g$ are continuous on the domain $D$. Let's compute the Jacobian determinants $\frac{\partial (y, z)}{\partial (x, y)}$, $\frac{\partial (z, x)}{\partial (x, y)}$ and $\frac{\partial (x, y)}{\partial (x, y)}$ required in our surface integral formula. We have that:

\begin{align} \quad \frac{\partial (y, z)}{\partial (x,y)} = \begin{vmatrix} \frac{\partial y}{\partial x} & \frac{\partial y}{\partial y}\\ \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y}\\ \end{vmatrix} = \begin{vmatrix} 0 & 1\\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{vmatrix} = - \frac{\partial g}{\partial x} \\ \quad \frac{\partial (z, x)}{\partial (x, y)} = \begin{vmatrix} \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} \\ \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y} \\ \end{vmatrix} = \begin{vmatrix} \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \\ 1 & 0 \end{vmatrix} = - \frac{\partial g}{\partial v} \\ \quad \frac{\partial (x, y)}{\partial (x, y)} = \begin{vmatrix} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y}\\ \frac{\partial y}{\partial x} & \frac{\partial y}{\partial y} \end{vmatrix} = \begin{vmatrix} 1 & 0\\ 0 & 1 \end{vmatrix} = 1 \end{align}

Therefore the surface integral of $w = f(x, y, z)$ over the surface $\delta$ where $\delta$ generated by $z = g(x, y)$ over $D$ on the $xy$-plane can be evaluated with the following formula:

\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} \: dx \: dy \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License