Computing Surface Area with Surface Integrals

# Computing Surface Area with Surface Integrals

Recall from the Surface Integrals page that if $\delta$ is a surface given parametrically by the equations $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ is given by the following formula:

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\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}

One nice feature of this formula is that we can compute the surface area of the surface $\delta$ given by the parametrically by $\vec{r}(t) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ to be:

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\begin{align} \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \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}

Let's look at some examples of finding surface areas.

## Example 1

Prove that the surface area of a sphere with radius $r$ is $4 \pi r^2$.

We first need to come up with a parameterization for a sphere. We can do this nicely by recalling from The Spherical Coordinate System page that we can use $\left\{\begin{matrix} x(\phi, \theta) = r \sin \phi \cos \theta\\ y(\phi, \theta) = r \sin \phi \sin \theta \\ z(\phi, \theta) = r cos \phi \end{matrix}\right.$, where $r$ is a fixed constant and for $0 ≤ \phi ≤ \pi$ and $0 ≤ \theta ≤ 2\pi$.

Now let's compute the necessary Jacobian determinants.

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\begin{align} \quad \frac{\partial (y, z)}{\partial (\phi, \theta)} = \begin{vmatrix} \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial \theta} \\ \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial \theta} \end{vmatrix} = \begin{vmatrix} r \cos \phi \sin \theta & r \sin \phi \cos \theta \\ -r \sin \phi & 0 \end{vmatrix} = r^2 \sin^2 \phi \cos \theta \end{align}
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\begin{align} \quad \frac{\partial (z, x)}{\partial (\phi, \theta)} = \begin{vmatrix} \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial \theta} \\ \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial \theta} \end{vmatrix} = \begin{vmatrix} -r \sin \phi & 0 \\ r \cos \phi \cos \theta & -r \sin \phi \sin \theta \end{vmatrix} = r^2 \sin^2 \phi \sin \theta \end{align}
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\begin{align} \quad \frac{\partial (x, y)}{\partial (\phi, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} r \cos \phi \cos \theta & -r \sin \phi \sin \theta \\ r \cos \phi \sin \theta & r \sin \phi \cos \theta \end{vmatrix} = r^2 \sin \phi \cos \phi \cos^2 \theta + r^2 \sin \phi \cos \phi \sin^2 \theta = r^2 \sin \phi \cos \phi \end{align}

Noting that $D = \{ (\phi, \theta) : 0 ≤ \phi ≤ \pi , 0 ≤ \theta ≤ 2\pi \}$ and that $\sqrt{r^4} = r^2$ since $r > 0$ and $\sqrt{\sin^2 \phi} = \sin \phi$ since $\sin \phi ≥ 0$ for $0 ≤ \phi ≤ \pi$ and we have that

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