Surface Integrals Examples 2

# Surface Integrals Examples 2

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 $\iint_{\delta} xz \: dS$ where $\delta$ is the surface generated by $z = x^2$ that lies inside the paraboloid $z = 1 - 3x^2 - y^2$ and in the first octant.

We should note that $\delta$ has a one-to-one projection onto the $xy$-plane. Let $G(x, y, z) = z - x^2 = 0$. Then we have that $\nabla G(x, y, z) = (-2x, 0, 1)$ and $\frac{\partial G}{\partial z} = 1$, so the area element $dS$ is given by:

(3)
\begin{align} \quad dS = \frac{\| \nabla G(x, y, z) \|}{\biggr \| \frac{\partial G}{\partial z} \biggr \|} \: dx \: dy = \frac{\sqrt{(-2x)^2 + (0)^2 + (1)^2}}{\sqrt{(1)^2}} \: dx \: dy = \sqrt{4x^2 + 1} \: dx \: dy \end{align}

Now since $z = x^2$ and $z = 1 - 3x^2 - y^2$ we have that:

(4)
\begin{align} \quad x^2 = 1 - 3x^2 - y^2 \\ \quad 4x^2 + y^2 = 1 \end{align}

Therefore the region $D$ is a quarter ellipse in the first quadrant of the $xy$-plane whose semi-major axis has length $1$ and whose semi-minor axis has length $\frac{1}{2}$. We can describe the region $D$ by:

(5)
\begin{align} \quad D = \{ (x, y) : 0 ≤ x ≤ \frac{1}{2}, 0 ≤ y ≤ \sqrt{1 - 4x^2} \} \end{align}

Therefore we have that:

(6)
\begin{align} \quad \iint_{\delta} xz \: dS = \iint_D x (x^2) \sqrt{4x^2 + 1} \: dx \: dy \\ \quad \iint_{\delta} xz \: dS = \int_0^{1/2} \int_0^{\sqrt{1 - 4x^2}} x^3 \sqrt{4x^2 + 1} \: dy \: dx \\ \quad \iint_{\delta} xz \: dS = \int_0^{1/2} x^3 \sqrt{4x^2 + 1} [y]_{y=0}^{y=\sqrt{1 - 4x^2}} \: dx \\ \quad \iint_{\delta} xz \: dS = \int_0^{1/2} x^3 \sqrt{(1 + 4x^2)(1 - 4x^2)} \: dx \\ \quad \iint_{\delta} xz \: dS = \int_0^{1/2} x^3 \sqrt{(1 - 16x^4} \: dx \end{align}

We rightmost single integral can be evaluated using substitution. Let $u = 1 - 16x^4$. Then $du = -64x^3 \: dx$ and so $-\frac{1}{64} du = x^3 \: dx$ and so:

(7)
\begin{align} \quad \int x^3 \sqrt{1 - 16x^4} \: dx = -\frac{1}{64} \int \sqrt{u} \: du = -\frac{2}{192} u^{3/2} = -\frac{1}{96} (1 - 16x^4)^{3/2} \end{align}

Therefore we have that:

(8)
\begin{align} \quad \iint_{\delta} xz \: dS = \left [ -\frac{1}{96} (1 - 16x^4)^{3/2} \right ]_{x=0}^{x=1/2} \\ \quad \iint_{\delta} xz \: dS = \frac{1}{96} \end{align}