Surface Integrals Examples 1

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 $\iint_{\delta} x \: dS$ where $\delta$ is the parabolic cylinder $z = \frac{x^2}{2}$ that is inside of the circular cylinder $x^2 + y^2 = 1$ in the first octant.

It should not be too difficult to see that the portion of $z = \frac{x^2}{2}$ cut out by the circular cylinder $x^2 + y^2 = 1$ in the first octant has a one-to-one projection onto the $xy$-plane.

Let $G(x, y, z) = z - \frac{x^2}{2} = 0$. Then $\nabla G(x, y, z) = (x, 0, 1)$ and $\frac{\partial G}{\partial z} = 1$. Also, let $D$ represent the quarter disk $x^2 + y^2 ≤ 1$ in the first quadrant of the $xy$-plane. Then we have that:

(3)

\begin{align} \quad \iint_{\delta} x \: dS = \iint_D x \frac{\| \nabla G (x, y, z) \|}{\biggr \| \frac{\partial G}{\partial z} \biggr \|} \: dx \: dy \\ \quad \iint_{\delta} x \: dS = \iint_D x \frac{\sqrt{(x)^2 + (0)^2 + (1)^2}}{\sqrt{(1)^2}} \: dx \: dy \\ \quad \iint_{\delta} x \: dS = \iint_D x \sqrt{x^2 + 1} \: dx \: dy \\ \quad \iint_{\delta} x \: dS = \int_0^1 \int_{0}^{\sqrt{1 - x^2}} x \sqrt{x^2 + 1} \: dy \: dx \\ \quad \iint_{\delta} x \: dS = \int_0^1 x\sqrt{x^2 + 1} \left [ y \right ]_{y = 0}^{y = \sqrt{1 - x^2}} \: dx \\ \quad \iint_{\delta} x \: dS = \int_0^1 x \sqrt{x^2 + 1} \sqrt{1 - x^2} \: dx \\ \quad \iint_{\delta} x \: dS = \int_0^1 x \sqrt{(1 + x^2)(1 - x^2)} \: dx \\ \quad \iint_{\delta} x \: dS = \int_0^1 x \sqrt{1 - x^4} \: dx \end{align}

Now we can evaluate the single integral on the righthand side with substitution. Let $u = x^2$. Then $du = 2x \: dx$ which implies that $\frac{1}{2} \: du = x \: dx$ and so:

(4)

\begin{align} \quad \int x \sqrt{1 - x^4} \: dx = \frac{1}{2} \int \sqrt{1 - u^2} \: du $]]. \end{align}

After noting applying the substitution to the limits of integration, notice that we then have the following indefinite integral $\quad \iint_{\delta} x \: dS = \frac{1}{2} \int_0^1 \sqrt{1 - u^2} \: du$. This integral is equal to half of the quarter area of the unit circle, that is, $\frac{1}{2} \int_0^1 \sqrt{1 - u^2} \: du = \frac{1}{2} \frac{\pi}{4} = \frac{\pi}{8}$.

Example 2

Evaluate $\iint_{\delta} y \: dS$ where $\delta$ is the portion of the plane $z = 1 + y$ that is inside the cone $z = \sqrt{2(x^2 + y^2)}$.

Once again, it's not hard to see that the plane $z = 1 + y$ has a one-to-one projection onto the $xy$-plane. Let $G(x, y, z) = z - 1 - y$. Then $\nabla G(x, y, z) = (0, -1, 1)$, and $\frac{\partial G}{\partial z} = 1$. Thus we have that:

(5)

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

Now note that since $z = 1 + y$ and $z = \sqrt{2(x^2 + y^2)}$ that in substituting the first equation into the second equation, we get that (by completing the square) the region $D$:

(6)

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

This is an equation of an ellipse centered at $(0, 1)$ with semi-minor axes having length $1$ and with semi-major axes having length $\sqrt{2}$. The area of this ellipse is therefore $\pi (1) (\sqrt{2}) = \pi \sqrt{2}$, and so:

(7)

\begin{align} \quad \iint_{\delta} y \: dS = \iint_D f(x, y, g(x, y)) \frac{\| \nabla G(x, y, z) \|}{\biggr \| \frac{\partial G}{\partial z} \biggr \|} \: dx \: dy \\ \quad \iint_{\delta} y \: dS = \sqrt{2} \iint_D y \: dx \: dy \\ \quad \iint_{\delta} y \: dS = \sqrt{2} (1) \sqrt{2} \pi \\ \quad \iint_{\delta} y \: dS = 2 \pi \end{align}