Flux through a surface
Before you start: you'll need unit normals from the gradient and double integrals.
\[\iint_S \vec F\cdot\hat n\,dS\]
where \(\hat n\) is the unit normal to the surface \(S\). It measures how much of the field flows through the surface.
Unit normal
For a surface \(\phi(x, y, z) = c\): \(\hat n = \dfrac{\nabla\phi}{|\nabla\phi|}\).
Projecting onto the \(xy\)-plane
\[dS = \frac{dx\,dy}{|\hat n\cdot\mathbf{k}|}\]
If water flows with velocity \(\vec F\), the flux is the volume of water crossing the surface per second. Only the part of \(\vec F\) along the normal counts; flow along the surface doesn't cross it.
The method
- Find \(\hat n\) from \(\nabla\phi\), choosing the direction the question asks for (usually outward or upward).
- Compute \(\vec F\cdot\hat n\), and use the surface equation to replace \(z\) (if projecting onto the \(xy\)-plane).
- Replace \(dS\) by \(\dfrac{dx\,dy}{|\hat n\cdot\mathbf{k}|}\) and integrate over the projection \(R\) of \(S\).
For spheres, it is often easier to use spherical coordinates, with \(dS = a^2\sin\theta\,d\theta\,d\phi\) on a sphere of radius \(a\).
Solved examples
Example 1: Flux through a plane
Exam levelEvaluate \(\displaystyle\iint_S \vec F\cdot\hat n\,dS\) for \(\vec F = 18z\,\mathbf{i} - 12\,\mathbf{j} + 3y\,\mathbf{k}\), where \(S\) is the part of the plane \(2x + 3y + 6z = 12\) in the first octant.
- \(\hat n = \dfrac{(2, 3, 6)}{7}\), and \(\hat n\cdot\mathbf{k} = \dfrac{6}{7}\), so \(dS = \dfrac{7}{6}\,dx\,dy\).
- \(\vec F\cdot\hat n = \dfrac{36z - 36 + 18y}{7}\). On the plane, \(6z = 12 - 2x - 3y\), so \(36z = 72 - 12x - 18y\) and \(\vec F\cdot\hat n = \dfrac{36 - 12x}{7}\).
- So the integral is \(\displaystyle\iint_R (6 - 2x)\,dx\,dy\) over the triangle \(2x + 3y \le 12\): \(\displaystyle\int_0^6 (6 - 2x)\,\frac{12 - 2x}{3}\,dx = \frac{1}{3}(432 - 648 + 288)\).
Answer\(24\)
Example 2: Through a sphere
EasyFind the flux of \(\vec r = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) out of the sphere \(x^2 + y^2 + z^2 = a^2\).
- On the sphere, \(\hat n = \dfrac{\vec r}{a}\), so \(\vec r\cdot\hat n = \dfrac{r^2}{a} = a\).
- Flux \(= a\times(\text{surface area}) = a\cdot 4\pi a^2\).
Answer\(4\pi a^3\)
Example 3: Through a hemisphere
MediumFind the flux of \(\vec F = z\,\mathbf{k}\) out of the upper hemisphere \(x^2 + y^2 + z^2 = a^2\), \(z \ge 0\).
- \(\hat n = \dfrac{\vec r}{a}\), so \(\vec F\cdot\hat n = \dfrac{z^2}{a}\). In spherical form, \(z = a\cos\theta\) and \(dS = a^2\sin\theta\,d\theta\,d\phi\).
- \(\displaystyle\int_0^{2\pi}\!\!\int_0^{\pi/2} a\cos^2\theta\cdot a^2\sin\theta\,d\theta\,d\phi = a^3\cdot\frac{1}{3}\cdot 2\pi\).
Answer\(\dfrac{2\pi a^3}{3}\)
Example 4: A scalar surface integral
EasyEvaluate \(\displaystyle\iint_S (x + y + z)\,dS\) over the part of \(x + y + z = 1\) in the first octant.
- On the plane the integrand is 1. \(\hat n = \dfrac{(1, 1, 1)}{\sqrt 3}\), so \(dS = \sqrt 3\,dx\,dy\).
- \(\sqrt 3\times\) (area of the triangle \(x + y \le 1\)) \(= \sqrt 3\cdot\dfrac{1}{2}\).
Answer\(\dfrac{\sqrt 3}{2}\)
Always substitute the surface equation into \(\vec F\cdot\hat n\) before integrating, as in Example 1. A leftover \(z\) in a double integral over \(x\) and \(y\) is a sure sign something was missed.
Common mistakes
1. Forgetting \(|\hat n\cdot\mathbf{k}|\)
\(dS\) is not \(dx\,dy\). A slanted surface has more area than its shadow.
2. Leaving \(z\) in the integrand
After projecting onto the \(xy\)-plane, eliminate \(z\) using the surface equation.
3. Choosing the wrong normal direction
Outward and inward normals give answers of opposite sign.
Practice questions
Q1Find \(\displaystyle\iint_S dS\) for the sphere of radius \(a\).
\(\displaystyle\int_0^{2\pi}\!\!\int_0^\pi a^2\sin\theta\,d\theta\,d\phi = 4\pi a^2\).
Q2Flux of \(\vec F = \mathbf{i}\) through the square \(x = 1\), \(0 \le y, z \le 1\), with \(\hat n = \mathbf{i}\).
\(\vec F\cdot\hat n = 1\), so the flux equals the area: 1.
Q3Flux of \(\vec r\) through the part of \(x + y + z = 1\) in the first octant (normal away from the origin).
\(\vec r\cdot\hat n = \dfrac{x + y + z}{\sqrt 3} = \dfrac{1}{\sqrt 3}\). Times the area \(\dfrac{\sqrt 3}{2}\): \(\dfrac{1}{2}\).
Q4Flux of \(\vec F = z\,\mathbf{k}\) upward through the disc \(z = 1\), \(x^2 + y^2 \le 4\).
\(\vec F\cdot\mathbf{k} = 1\) on the disc, so the flux is its area, \(4\pi\).
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Frequently asked questions
What does a surface integral of a vector field measure?
The flux: the net amount of the field passing through the surface, counting only the component along the normal.
How do I find dS?
Project onto a coordinate plane: \(dS = \dfrac{dx\,dy}{|\hat n\cdot\mathbf{k}|}\), or use \(a^2\sin\theta\,d\theta\,d\phi\) on a sphere of radius \(a\).
Where surface integrals lead next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.