Surface Integrals Examples 5
 Table of Contents

Surface Integrals Examples 5

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 $\iint_{\delta} y \: dS$ where $\delta$ is the portion of the paraboloid $y = x^2 + z^2$ that is contained in the cylinder $x^2 + z^2 = 4$.

The surface $\delta$ is shown below:

We note that the surface $\delta$ has a one-to-one projection onto the $xz$-plane. Let $G(x, y, z) = y - x^2 - z^2 = 0$. The surface area element $dS$ is given by:

(3)
\begin{align} \quad dS = \frac{\nabla G(x, y, z)}{\frac{\partial G}{\partial y}} \: dx \: dz \end{align}

We note that $\nabla G(x, y, z) = (-2x, 1, -2z)$ and $\frac{\partial G}{\partial y} = 1$. Therefore:

(4)
\begin{align} \quad dS = \sqrt{(-2x)^2 + (1)^2 + (-2z)^2} \: dx \: dz = \sqrt{4x^2 + 1 + 4z^2} \end{align}

Now $D$ will be the projection of $\delta$ onto the $xz$-plane which will be a circle of radius $2$ centered at the origin $(0, 0)$ on the $xz$-plane. Let $x = r \cos \theta$ and $z = r \sin \theta$. In polar coordinates we have that $D$ can be expressed as:

(5)
\begin{align} \quad D = \{ (r, \theta) : 0 ≤ r ≤ 2 : 0 ≤ \theta ≤ 2\pi \} \end{align}

Thus using polar coordinates to evaluate our surface integral and noting that $y = x^2 + z^2 = r^2$ we have that:

(6)
\begin{align} \quad \iint_{\delta} y \: dS = \iint_D r^2 \sqrt{4r^2 + 1} r \: dr \: d \theta \\ \quad \iint_{\delta} y \: dS = \int_0^{2\pi} \int_0^2 r^2 \sqrt{4r^2 + 1} r \: dr \: d \theta \end{align}

We can evaluate the rightmost inner integral using substitution. Let $u = 4r^2 + 1$. Then $\frac{u - 1}{4} = r^2$. We also have that $du = 8r \: dr$ so $\frac{1}{8} du = r \: dr$ and so:

(7)
\begin{align} \quad \int r^2 \sqrt{4r^2 + 1} r \: dr = \frac{1}{8} \int \frac{u - 1}{4} \sqrt{u} \: du = \frac{1}{32} \int (u - 1) \sqrt{u} \: du = \frac{1}{32} \int (u^{3/2} - \sqrt{u}) \: du \\ = \frac{1}{32} \left ( \frac{2}{5} u^{5/2} - \frac{1}{2}u^2 \right ) = \frac{1}{32} \left ( \frac{2}{5} (4r^2 + 1)^{5/2} - \frac{2}{3}(4r^2 + 1)^{3/2} \right ) \end{align}

Therefore we have that:

(8)
\begin{align} \quad \iint_{\delta} y \: dS = \int_0^{2\pi} \left [ \frac{1}{32} \left ( \frac{2}{5} (4r^2 + 1)^{5/2} - \frac{2}{3}(4r^2 + 1)^{3/2} \right ) \right ]_{r=0}^{r=2} \: d \theta \\ \quad \iint_{\delta} y \: dS = \int_0^{2\pi} \left [ \frac{1}{32} \left ( \frac{2}{5} (17)^{5/2} - \frac{2}{3} 17^{3/2} \right ) - \left ( \frac{2}{5} - \frac{2}{3} \right ) \right ] \: d \theta \\ \quad \iint_{\delta} y \: dS = \left [ \frac{1}{32} \left ( \frac{2}{5} (17)^{5/2} - \frac{2}{3} 17^{3/2} \right ) - \left ( \frac{2}{5} - \frac{2}{3} \right ) \right ] \int_0^{2\pi} \: d \theta \\ \quad \iint_{\delta} y \: dS =\frac{\pi}{16} \left ( \frac{2(17^{5/2} - 1)}{5} - \frac{2(17^{3/2} - 1)}{3} \right ) \end{align}
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