A surface integral adds up a quantity over a surface, in the same way that an ordinary integral adds up a quantity along a line. The surface can be flat or curved, open or closed. Two kinds appear often in engineering. The first adds a scalar over the surface, and the total surface area is the simplest case of it. The second adds the normal component of a vector field, and this total is called the flux. This page shows both kinds with a worked example each. Then it gives the general recipe that works for any surface you can describe with two parameters.

If you are new to the concept of Integration, I would suggest you to read the "Integration" page first.

The topics on this page are listed below.

• What does a surface integral add up ?
• Examples 01 - Gauss Law
• Examples 02 - Surface Area of a Sphere
• How do you compute a surface integral on a general surface ?

What does a surface integral add up ?

Let's start from a picture before any formula. We cut the surface into many small patches. At every patch we evaluate something, and then we add all the results. The only question is what we evaluate at each patch, and the example below answers it for the vector case.

As you may guess from the word itself, surface integral is a type of integraion taken over a surface. There can be many different operations over a surface. One example is as shown below.

In this example, you see two vectors in each segments of the surface. one of the vector is normal to each surface segment(This vector is called 'normal vector').  The other vector is an arbitrary angle to the normal vector . Now I want to take the inner product of each red vector and blue vector and sum them all. This operation can be represented in a mathematical form as shown below. The mathematical operation is exactly same as the one shown in previous example. The only differences is that this operation goes along the surface. This kind of integration is called "Surface Integral".

The picture has two parts. The upper part is a curved surface with many small arrows on it. The lower part magnifies one patch and writes the integral under it.

• The red arrow on the magnified patch is ds. It is normal to the patch, and its length stands for the area of the patch.
• The blue arrow is the field vector F at the same patch. It can point in any direction relative to the patch.
• The inner product F ⋅ ds keeps only the part of F that crosses the patch. A field that runs along the surface gives zero, and a field that crosses it at a right angle gives the full |F| |ds|.
• The double integral sign says that the patches cover a two dimensional region. So the sum runs over two directions on the surface, not along one line.

The "previous example" in the paragraph above is the line integral, where the same inner product is taken along a curve. You can find it on the Line Integral page. The surface version changes only the region of the sum. It replaces a small step along a curve with a small patch of a surface.

Followings are some of the exaples that uses the surface integral and I will keep adding more examples as I have chance.  These examples would not be a completely new one. You would find these examples from the various textbook, but I would explain it in my own way and hopefully help you to understand more easily and intuitively.

• Example 1 > Gauss Law
• Example 2 > Surface Area of a Sphere

• A surface integral is a sum over small patches : each patch contributes one small value, and the integral adds them over the whole surface.
• The flux uses the normal component only : the inner product with the normal vector drops the part of the field that runs along the surface.
• A scalar surface integral has no inner product : it adds a plain number times the patch area, and the total area is the case where that number is 1.

Examples 01 - Gauss Law

Gauss Law is the flux integral in its most famous form. It connects the flux of the electric field through a closed surface to the charge enclosed by that surface. So it is a good first example of the vector kind of surface integral.

One of the most common example of surface integral is Gauss Law of electric field which is expressed as shown below.

(This is one component of Maxwell equations in electromagnetic theory).

Meaning of the left hand part of the equation (Integration Part) is as shown below. Suppose that you have a closed surface (like a sphere) and put a charge inside of the closed surface. Then the electric field line goes out of the charge (if the charge is positive) and penerate out of the closed surface. If you take a small segment (rectangular plane) of the surface, you will see the field line going through the surface with a certain angle to the plane. The angle between the field line and the surface is defined by the inner product of the normal vector and the field line. If you take the inner product of the two vectors of each segnment accorss the whole surface, it give you the value of the left hand side of the equation.

In the picture, the blue arrow E is the electric field at one patch and the red arrow dA is the normal of that patch. The note beside dA says that its magnitude represents the area of the patch. The oval on the left is a closed surface around a charge at its centre. To be precise about the angle, the inner product gives E ⋅ dA = |E| |dA| cos θ, where θ is the angle between the field and the normal vector. So a field that crosses the patch at a right angle, parallel to the normal, gives the largest contribution.

Let's check the law with the simplest case. A point charge Q sits at the centre of a sphere of radius r. By symmetry, the field on the sphere points straight outward and has the same size everywhere, E = Q / (4 π ε₀ r²). The field is then parallel to dA at every patch, so E ⋅ dA is just E dA. E is constant over the sphere, so it comes out of the integral. What remains is the total area of the sphere. Examples 02 below shows that this area is 4 π r². The flux is therefore Q / (4 π ε₀ r²) x 4 π r² = Q / ε₀, which is exactly the right hand side of the law.

Notice that r cancels. The flux does not depend on the size of the sphere. It does not depend on the shape of the surface either, as long as the surface encloses the same charge. This is why Gauss Law is useful. You can choose the surface that makes the integral easy.

• The left side of Gauss Law is a flux integral : it adds E ⋅ dA over every patch of a closed surface.
• Symmetry turns the flux into field times area : when E is normal to the surface and constant on it, the integral for a sphere is E x 4 π r².
• The result depends only on the enclosed charge : the radius cancels, so every closed surface around Q gives the same flux Q / ε₀.

Examples 02 - Surface Area of a Sphere

This example is the scalar kind of surface integral, with the number 1 at every patch. The hard part is not the integration itself. It is writing the area of one small patch in terms of the two angles that locate it on the sphere.

This example itself would be one of the most common one you may see in your multi-variable calculus textbook. But I want to explain it in a little bit different way from what you see in the textbook (hopefully my explanation would help you better to understand this example).

The question given to you in this example is 'what is the total surface area of the sphere shown below ?'.

If you have any basic concept of integration and understand you can apply the concept to solve this problem, you may easily come up with the idea that 'if I split the shole surface into a lot of small area and sum up all of those small area across the whole surface, you can get the total area of the surface'.

The first step is to turn this into mathematical form. When you do this, I would suggest you not to worry too much of how to solve it. If you think too much of getting the solution from the first place, you would have more trouble reaching to the solution.

My first mathematical interpretation for the satement 'split the shole surface into a lot of small area and sum up all of those small area across the whole surface' is as follows. Isn't it simple ?

Or you may write it as follows.

The meaning of this expression can be illustrated as below. I hope this illustration and short comments on it is good enough for your understanding.

The expression written above is very simple and not so much scary, it does not have much details on getting the specific solution. In order to bring you closer to the specific solution, I would break down the term dA into more detailed parameters as shown below. If you assume that the red shape is a rectangle, the area (dA) can be calculated by (dx * dy). In this illustration, the red area would not look like a rectangle (it would look like a trapezoid), however if you break down the area very small you can approximate it as a rectangle.

Now let's rewrite the equation by replacing the term dA with dy dx. It becomes as follows. Since the single variable dA has become two variable (dx dy), the number of integral symbol becomes two as well. Getting scared ? Don't worry.. I would not solve this equation for now.. just try to understand the meaning of the equation.

Now the next step is to figure out how to figure out the dx and dy. In order to do this, I would need to expand the picture and put many lines and labels as shown below. However, you would need to put your own effort on this to clearly understand the meaning of each labels and lines. It is difficult to show you the 3 dimensional object into 2 dimensional surface (the monitor). Actualy even if you see this in real 3D space, it would not be so easy unless you put some of your own effort.

Using the additional lines and parameters (i.e, radius, angles), you can figure out dx and dy as shown below. In this figure, dx and dy is calculated by the arc length. As you know, an arc is not the straight line.. but if you make the red area small enough you can approximate the arc as a straight line that represents a side of a rectangle.

From the high school math (junior high school or senior high school depending on where you are educated), you would remember that you can calculate the length of an arc on the circumference of a circle from the radius and angle.(If you arleady forgot about this, like me :), refer to the definition of arc length page in www.mathsisfun.com ). Using this arc length calculation method, you can figure out dx and dy as shown below. It would take some time and effort for you to understand how dx and dy can be written as follows. But this is the most important step. So make it sure that you clearly understand this figure before you move any further.

The two arc lengths in the picture above come from two different circles. The green circle is a meridian through the poles, and its radius is the full r. So a step dφ along it has length r dφ. The black circle is a circle of latitude. Its centre sits on the vertical axis, and its radius is only r sin(φ), because φ is measured from the vertical axis. So a step dθ around it has length r sin(φ) dθ. This is the reason the patches shrink near the poles, where sin(φ) goes to 0.

With this new expression, you can label the dx and dy in the equation as shown below.

If you repalce the term dx and dy with the new information you get, the equation should become as below.

Now let's specify the range of the two angles. I don't have any good idea to show this range clearly in two dimentional drawing. Just for the conclusion the range of the two angle is specified as follows. Think of how this range came out on your own. You may not get it just in a couple of seconds.. but if you keep thinking I think you can figure it out in several minutes. Specifying the range of the angle and with a little bit of rewriting, you would get following equation.

Simplifying the equation just a little bit, you can rewrite it as shown below.

In case of sphere, the radius r is same in every point on the surface. it means the r is a constant. So you can pull out the r term out of the integration as below. (If the shape of the object is not a sphere where r changes depending on location, just put the r as shown above).

Let's finish the calculation, because the equation above still has the integral signs in it. The inner integral is ∫ sin(θ) dθ from 0 to π. Its antiderivative is -cos(θ), so the value is -cos(π) + cos(0) = 1 + 1 = 2. The outer integral adds 2 over φ from 0 to 2π, which gives 2 x 2π = 4π. The total is therefore A = 4 π r². For r = 1 this is about 12.566, the well known area of a unit sphere.

You may notice that the two angles change their roles from one picture to the next. In the drawings of the patch, φ is measured from the vertical axis and θ runs around it. So the patch there is r sin(φ) dθ x r dφ. From the equation with the two angles onward, the symbols are swapped. There θ is the angle from the vertical axis, and φ runs around it. The picture that labels dy and dx also points each arrow at the opposite factor from the picture before it. Neither swap changes the answer, because the product of the two sides is the same. What matters is the pairing. The sine always belongs to the angle measured from the vertical axis, and that angle runs from 0 to π. The angle around the axis runs from 0 to 2π. The equations from the ranges onward follow this pairing, with θ as the angle from the vertical axis.

The ranges also have a simple reason. The angle from the vertical axis goes from the north pole, at 0, to the south pole, at π. One full turn around the axis is 2π. Together the two ranges cover every point of the sphere exactly once.

• The patch area on a sphere is r² sin(θ) dθ dφ : here θ is the angle from the vertical axis, and the sine appears because circles of latitude get smaller near the poles.
• The polar angle runs from 0 to π, the azimuth from 0 to 2π : with these ranges every point of the sphere is counted once.
• The integral gives 4 π r² : the sine integral contributes 2 and the full turn contributes 2π.
• Check which angle carries the sine : the drawings and the equations on this page use θ and φ for different angles, but the sine always goes with the angle from the vertical axis.

How do you compute a surface integral on a general surface ?

The sphere worked because we could find the patch area with arc lengths. Most surfaces do not allow that. So we need a recipe that gives the patch area, and the normal vector, for any surface we can describe with two parameters.

Let's describe the surface with a position vector r(u, v). The two parameters u and v play the role of the two angles on the sphere. If we change u by du, the point moves by r_u du, where r_u is the partial derivative of r with respect to u. If we change v by dv, the point moves by r_v dv. These two small steps are the two sides of the patch. The patch is a small parallelogram, not always a rectangle.

The cross product of the two sides gives everything we need. Call it N = r_u x r_v, where x here means the cross product. Its direction is normal to the surface. Its length |N| is the area of the parallelogram per unit du dv. So the two kinds of surface integral become ordinary double integrals over u and v.

• Scalar surface integral : ∫∫ f dS = ∫∫ f(r(u, v)) |N| du dv. With f = 1 this is the area.
• Flux of a vector field : ∫∫ F ⋅ dS = ∫∫ F(r(u, v)) ⋅ N du dv. Here the length of N already carries the area, so no extra factor is needed.

Let's check the recipe against the sphere. Take r(θ, φ) = (R sinθ cosφ, R sinθ sinφ, R cosθ), with θ from the vertical axis. Working out the cross product gives |N| = R² sinθ. This is exactly the r² sin(θ) of Examples 02, found this time without any drawing.

The same parametrization also gives a flux. Take the field F = (x, y, z), which points straight outward from the origin. On the sphere, F ⋅ N = R³ sinθ. Integrating over θ from 0 to π and φ from 0 to 2π gives a flux of 4 π R³. The divergence gives the same number from the inside. The divergence of this field is 3 everywhere, and 3 times the volume 4/3 π R³ is again 4 π R³. This agreement is the divergence theorem, and Gauss Law is the same theorem applied to the electric field.

A surface given as a graph z = f(x, y) is a common special case. Use x and y as the two parameters. Then |N| = √(1 + f_x² + f_y²), and the area element is this factor times dx dy. For example, the paraboloid z = x² + y² over the unit disk has an area of π/6 x (5√5 - 1), which is about 5.330. The flat disk under it has an area of only π, about 3.142. The difference comes from the tilt of the surface, and the square root factor measures that tilt.

• Two parameters describe the surface : the position vector r(u, v) plays the role that the two angles play on the sphere.
• The cross product of the two tangent vectors does two jobs : its direction is the normal, and its length is the patch area per du dv.
• The recipe reproduces the sphere result : for the sphere it gives R² sinθ, the same factor that Examples 02 found from arc lengths.
• A flux through a closed surface can be checked with the divergence : for F = (x, y, z) both ways give 4 π R³.