Surface Integral in Multivariable Calculus

What Does Integral Mean?

To understand surface integral, it is very important to understand what an integral means. Integration and differentiation are the two sides of calculus. The two basic operations, where on one hand differentiation helps us to examine the rates of change and on the other hand integration helps us to add up infinitesimal pieces of a whole. It will be fruitless to understand integration without an example. So here is an irregular shape divided into regular rectangles of whose area integral is what we desire to calculate.

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If we add up the area of all the regular rectangles, we can find the area integral of the irregular shape. If we keep making the rectangles thinner and thinner, our approximation of the area integral of the whole irregular shape would become more and more accurate. And finally when the rectangles will be infinitely small, our area integral of the shape would be perfect. This is what an integral does. It allows us to find the area, volumes, central points and displacement. Now that we know what an integration means, let us not delay in digging what a surface integration means.

Surface Integral Meaning

A surface integral is just like a line integral. In a line integral , we integrate over a path in a plane which is one dimensional and on the surface integral, we integrate over a surface which can be two dimensional or three dimensional. Therefore, we need a new kind of integral that can integrate over objects in higher dimensions and for which we need to expand the concept from a line integral to a surface integral. 

We definitely want to know how a surface integral can be calculated and what is it used for and also a surface integral example but first we would want to know how they are defined. Therefore, the surface integral of the function f(x, y, z) over the surface S will be denoted by

                                       ∫∫_sf(x,y,z)ds ……(1)

Now, dS is considered as the area of an infinitesimal piece of the surface S. in order to define the integral (1), we have to subdivide the surface S into small pieces having area ∆Si then  pick a point (xi , yi , zi) in the i-th piece, to form the Riemann sum.

                                       ∑f(xᵢ,yᵢ,zᵢ)ΔSᵢ ……(2)    

As the  subdivision of S gets finer, the corresponding sums (2) will reach a limit where it will not be dependent on the choice of the points or how the surface was subdivided. Here, the surface integral (1) is defined to be this limit. (The surface must be smooth and not infinite in extent. Also, the subdivisions must be made reasonably or else either the limit may not exist, or it may not be unique.)

Parametric Surfaces

Before we integrate over a surface and surface integral example, we must first consider the surface itself. If we recall, we will remember that in order to calculate a scalar or vector line integral over curve C, we had to first parameterize C. Therefore, to calculate a surface integral over surface S, we have to parameterize S. And to do that, we require a working concept of a parameterized surface just like we already had a concept of a parameterized curve.

A parameterized surface can be represented as:

r(u,v) = (x(u,v), y(u,v), z(u,v))       

We can see that the parameterization includes two parameters, u and v, that is because the surface is two-dimensional, and therefore we need two variables to trace out the surface. The parameters u and v vary over a region which can be called the parameter domain. The set of points in the UV-plane can be substituted into r. Each choice of u and v in the parameter domain produces a point on the surface, just like each choice of a parameter t gives a point on a parameterized curve. The whole surface is created by making possible choices of u and v over the parameter domain. Just like line integrals, surface integrals are of two kinds:

1. A surface integral of a scalar-valued function.

2. A surface integral of a vector field.

Surface Integral of a Scalar-Valued Function 

Now that we are able to parameterize surfaces and calculate their surface areas, we are ready to define surface integrals. We can start with the surface integral of a scalar-valued function. Now it is time for a surface integral example:

Consider a surface S and its function f(x, y, z)

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If S is denoted by the position vector, r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k, then the surface integral of the scalar function would be:

 ∫∫_sf(x,y,z)dS =  ∫∫_D(u,v)f[x(u,v), y(u,v), z(u,v)] ⏐\[\frac{∂r}{∂u}\] x \[\frac{∂r}{∂v}\]⏐dudv

Where, the range of coordinates over the domain of the UV-plane are (u,v)

The cross product which is perpendicular to the surface at a point (u,v) are \[\frac{∂r}{∂u}\] x \[\frac{∂r}{∂v}\]

The partial derivatives are \[\frac{∂r}{∂u}\] and \[\frac{∂r}{∂v}\]

Therefore, the absolute value  ⏐\[\frac{∂r}{∂u}\] x \[\frac{∂r}{∂v}\]⏐ can be referred to as the area element.

Surface Integrals of Vector Fields

To calculate the surface integrals of vector fields, consider a vector field with surface S and function F(x,y,z). It is continuously defined by the vector position r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k.

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Now let n(x,y,z) be a normal vector unit to the surface S at the point (x,y,z). Here the surface S is smooth and has a continuous function n(x,y,z). There are two possibilities such as n(x,y,z) and n(x,y,z). 

If we make a choice, the surface will be oriented which can either be inward or outward. 

a) When the Surface S is Oriented is Outward:

 ∫∫_sF(x,y,z).dS = ∫∫_sF(x,y,z).ndS =  ∫∫_D(u,v)F[x(u,v), y(u,v), z(u,v)] . ⏐\[\frac{∂r}{∂u}\] x \[\frac{∂r}{∂v}\]⏐dudv

b) When the Surface S is Oriented is Inward:

 ∫∫_sF(x,y,z).dS =  ∫∫_sF(x,y,z). ndS =  ∫∫_D(u,v)F[x(u,v), y(u,v), z(u,v)] . ⏐\[\frac{∂r}{∂u}\] x \[\frac{∂r}{∂v}\]⏐dudv

dS = ndS = vector element of the surface. s

Solved Examples

Question 1) The equations z = 12, x² + y² ≤ 25

Describe the disk having radius 5 that lie on the plane z = 12. Consider that r is the position vector field r(x, y, z) = xi + yj + zk.

Therefore, Compute,

                                            ∫∫_sr.dS.

Solution 1) Now since the disk is very much parallel to the xy plane, the outward unit normal is k. Hence n(x, y, z) = k and so r · n = z. Thus,

∫∫_sr.dS = ∫∫_sr.ndS = ∫∫_sr.zdS = ∫∫_D12dx dy = 300π

There is an alternative that is we can solve this problem with the help of the formula for surface integrals over graphs: 

∫∫_sF.dS = ∫∫_DF(-\[\frac{∂g}{∂x}\]i - \[\frac{∂g}{∂y}\]j + k)dx dy.

With g(x, y) = 12 and D the disk x² + y² ≤ 25, we will get

∫∫_sr.dS = ∫∫_D(x.0 + y.0 + 12)dx dy = 12(area of D) = 300π

Question 2) Evaluate the surface integral of the vector field F = 3x²i − 2yxj + 8k over the surface S that is the graph of z = 2x − y over the rectangle [0, 2] × [0, 2].

Solution 2) We will use the formula of the surface integral over a graph z = g(x, y) :

∫∫_sF.dS =  ∫∫_D[F . (-\[\frac{∂g}{∂x}\]i - \[\frac{∂g}{∂y}\]j + k)dx dy.

In this case, we will get: 

∫\[_{0}^{2}\]∫\[_{0}^{2}\](3x²-2yx,8.) . (-2,1,1)dx dy∫\[_{0}^{2}\]∫\[_{0}^{2}\](-6x²,-2yx,+8)dx dy

∫\[_{0}^{2}\]- 2x³ - yx² + 8x 

∫\[_{0}^{2}\] - 4ydy = -2y² l\[_{0}^{2}\] = -8.