Surface Integrals

The Essentials

A surface integral is calculated by integrating over all of the little pieces of surface. A little piece of surface is:

\[ \vec{t}_u\times \vec{t}_v \]

Where \( \vec{t}_u \) is the tangent vector in the u direction \( \langle x_u,y_u,z_u\rangle \), and \( \vec{t}_v \) is the tangent vector in the v direction \( \langle x_v,y_v,z_v\rangle \) where u and v come from the parameterization of the surface \( r(u,v) \). This is true because the magnitude of the cross product of two vectors is the area of the parallelogram that the two vectors make. Therefore, the surface area of a surface given a parameterization is:

\[ \iint_D ||\vec{t}_u\times\vec{t}_v|| dA \]

The formula for the surface integral of a scalar function is:

\[ \iint_Sf(x,y,z)dS=\iint_D f(\vec{r}(u,v)) ||\vec{t}_u\times\vec{t}_v|| dA \]

The unit normal vector is:

\[ \vec{N}=\frac{\vec{t}_u\times\vec{t}_v}{||\vec{t}_u\times\vec{t}_v||} \]

The formula for the vector surface integral is:

\[ \iint_S\vec{F}\cdot d\vec{S}=\iint_S\vec{F}\cdot \vec{N}dS=\iint_D \vec{F}(\vec{r}(u,v))\cdot(\vec{t}_u\times\vec{t}_v)dA \]

A vector surface integral is also called a flux integral.

Example

Calculate the surface integral of the function

\[ \vec{F}=\langle -xz, -yz, z^2\rangle \]

over the surface parameterized by the function

\[ \vec{r}=\langle u\cos(v),u\sin(v),u\rangle \]

from \( 0\leq u\leq 4 \) and \( 0\leq v\leq 2\pi \).

The function is a vector function, so we will use the vector function formula. To find the \( \vec{t}_u\times\vec{t}_v \):

\[ \vec{r'}_u=\vec{t}_u=\langle \cos(v),\sin(v),1\rangle\hspace{36pt}\vec{r'}_v=\vec{t}_v=\langle -u\sin(v),u\cos(v),0\rangle \]
\[ \vec{t}_u\times\vec{t}_v=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \cos(v) & \sin(v) & 1\\ -u\sin(v) & u\cos(v) & 0 \end{vmatrix}=\langle -u\cos(v),-u\sin(v),u\rangle \]

Now to calculate the integral:

\[ \iint_D \vec{F}(\vec{r}(u,v))\cdot(\vec{t}_u\times\vec{t}_v)dA \]
\[ \int_0^{2\pi}\int_0^4 \langle -u^2\cos(v),-u^2\sin(v),u^2\rangle\cdot\langle -u\cos(v),-u\sin(v),u\rangle dudv \]
\[ \int_0^{2\pi}\int_0^4 u^3\cos^2(v)+u^3\sin^2(v)+u^3dudv \]
\[ \int_0^{2\pi}2 dv\int_0^4 u^3du \]
\[ 4\pi(64) \]
\[ 256\pi \]

Practice

  1. Evaluate the surface integral over the top half of the surface of the sphere with a radius of 3 centered on the origin:
    \[ \iint_D xz dA \]
  2. Evaluate the surface integral of the function \( \langle -x,3y,z\rangle \) over the surface defined by the plane \( 2z+2y-x=5 \) from \( 0\leq x\leq 1 \) and \( 0\leq y\leq 1 \)

Solutions:

  1. 0
  2. 4

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