There are several equations for surface integrals and which one you use depends on what form your equations are in. This is a similar situation that you encountered with line integrals.
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There are two main groups of equations, one for surface integrals of scalarvalued functions and a second group for surface integrals of vector fields (often called flux integrals). The following table places them sidebyside so that you can easily see the difference.
scalarvalued function 
\(f(x,y,z)\)  
vector field 
\(\vec{F}(x,y,z)=M(x,y,z)\hat{i} + N(x,y,z)\hat{j} + P(x,y,z)\hat{k}\) 
What Are Surface Integrals?
Before we get started with the details of surface integrals and how to evaluate them, let's watch a couple of great videos that will gently introduce you to surface integrals of scalarvalued functions and give you some examples. This is one of our favorite instructors and we think these videos are worth taking the time to watch.
video by Dr Chris Tisdell 

In this second video from about the 15min20sec mark to the end of the video, he discusses a way to simplify the integration in a special case. We suggest that you do not watch that part of the video until you have worked some practice problems and are comfortable with the basics of setting up the integrals. However, check with your instructor to see if they will allow you to use this technique.
video by Dr Chris Tisdell 

Surface Integrals of ScalarValued Functions
The following equations are used when you are given a scalarvalued function over which you need to evaluate a surface integral. The form of the function is \(f(x,y,z)\). In general, your surface is parameterized as \(\vec{r}(u,v)=x(u,v)\hat{i} + y(u,v)\hat{j} + z(u,v)\hat{k}\). So to evaluate the integral of \(f(x,y,z)\) over the surface \(\vec{r}\), we use the equation
\( \iint\limits_S {f(x,y) ~ dS} = \iint\limits_R {f(x(u,v),y(u,v),z(u,v)) ~ \ \vec{r}_u \times \vec{r}_v \ ~ dA} \) 
\(\vec{r}(u,v)\) is the parametric surface 
R is the region in the uvplane 
\(\vec{r}_u\) and \(\vec{r}_v\) are the partial derivatives of \(\vec{r}\) 
In the special case where we have the surface described as \(z=g(x,y)\), we can parameterize the surface as \(\vec{r}=x\hat{i}+y\hat{j}+g(x,y)\hat{k}\). This gives \( \ \vec{r}_x \times \vec{r}_y \ = \sqrt{1+[g_x]^2+[g_y]^2} \) and the surface integral can then be written as \( \iint\limits_S {f(x,y) ~ dS} = \iint\limits_R {f(x,y,z) ~ \sqrt{1+[g_x]^2+[g_y]^2} ~ dA} \) and R is the region in the xyplane.
Okay, let's try some practice problems evaluating surface integrals of scalar functions.
Practice  Surface Integrals, ScalarValued Functions
Basic 

Evaluate \( \iint_S { x^2 y z ~ dS } \) where S is the part of the plane \( z = 1 + 2x + 3y \) that lies above the rectangle \( 0 \leq x \leq 3, 0 \leq y \leq 2 \).
Problem Statement 

Evaluate \( \iint_S { x^2 y z ~ dS } \) where S is the part of the plane \( z = 1 + 2x + 3y \) that lies above the rectangle \( 0 \leq x \leq 3, 0 \leq y \leq 2 \).
Final Answer 

\( 171 \sqrt{14} \)
Problem Statement
Evaluate \( \iint_S { x^2 y z ~ dS } \) where S is the part of the plane \( z = 1 + 2x + 3y \) that lies above the rectangle \( 0 \leq x \leq 3, 0 \leq y \leq 2 \).
Solution
video by PatrickJMT 

Final Answer
\( 171 \sqrt{14} \)
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Evaluate \( \iint_S{ xy ~ dS } \) using a parametric surface where S is \( x^2 + y^2 = 4, 0 \leq z \leq 8 \) in the first octant.
Problem Statement 

Evaluate \( \iint_S{ xy ~ dS } \) using a parametric surface where S is \( x^2 + y^2 = 4, 0 \leq z \leq 8 \) in the first octant.
Final Answer 

32
Problem Statement
Evaluate \( \iint_S{ xy ~ dS } \) using a parametric surface where S is \( x^2 + y^2 = 4, 0 \leq z \leq 8 \) in the first octant.
Solution
video by MIP4U 

Final Answer
32
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Evaluate \( \iint_S { x^2 + y^2 } \) using a parametric surface where S is the hemisphere \( x^2 + y^2 + z^2 = 1 \) above the xyplane.
Problem Statement 

Evaluate \( \iint_S { x^2 + y^2 } \) using a parametric surface where S is the hemisphere \( x^2 + y^2 + z^2 = 1 \) above the xyplane.
Final Answer 

\(4\pi/3\)
Problem Statement
Evaluate \( \iint_S { x^2 + y^2 } \) using a parametric surface where S is the hemisphere \( x^2 + y^2 + z^2 = 1 \) above the xyplane.
Solution
video by MIP4U 

Final Answer
\(4\pi/3\)
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Integrate \( f(x,y,z) = xy \) over the surface \( z = 4  2x  2y \) in the first octant.
Problem Statement 

Integrate \( f(x,y,z) = xy \) over the surface \( z = 4  2x  2y \) in the first octant.
Final Answer 

2
Problem Statement
Integrate \( f(x,y,z) = xy \) over the surface \( z = 4  2x  2y \) in the first octant.
Solution
video by MIP4U 

Final Answer
2
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Intermediate 

Compute the surface integral of \(\displaystyle{ f(x,y,z) = \frac{2z^2}{x^2+y^2+z^2} }\) over the cap of the sphere \( x^2 + y^2 + z^2 = 9 \), \(z \geq 2\).
Problem Statement 

Compute the surface integral of \(\displaystyle{ f(x,y,z) = \frac{2z^2}{x^2+y^2+z^2} }\) over the cap of the sphere \( x^2 + y^2 + z^2 = 9 \), \(z \geq 2\).
Final Answer 

\( 76 \pi/9 \)
Problem Statement
Compute the surface integral of \(\displaystyle{ f(x,y,z) = \frac{2z^2}{x^2+y^2+z^2} }\) over the cap of the sphere \( x^2 + y^2 + z^2 = 9 \), \(z \geq 2\).
Solution
Note  At about the 13min50sec mark, he writes \(76\pi/3\) as the answer. At the end of the video, he corrects his answer to \(76\pi/9\).
video by Dr Chris Tisdell 

Final Answer
\( 76 \pi/9 \)
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A roof is given by the graph of \( g(x,y) = 25 + 0.5x + 0.5y \) over \( 0 \leq x \leq 40 \), \( 0 \leq y \leq 20 \). If the density of the roof is given by \( f(x,y,z) = 150  2z \), determine the mass of the roof.
Problem Statement 

A roof is given by the graph of \( g(x,y) = 25 + 0.5x + 0.5y \) over \( 0 \leq x \leq 40 \), \( 0 \leq y \leq 20 \). If the density of the roof is given by \( f(x,y,z) = 150  2z \), determine the mass of the roof.
Final Answer 

\( 281994\sqrt{6} \)
Problem Statement
A roof is given by the graph of \( g(x,y) = 25 + 0.5x + 0.5y \) over \( 0 \leq x \leq 40 \), \( 0 \leq y \leq 20 \). If the density of the roof is given by \( f(x,y,z) = 150  2z \), determine the mass of the roof.
Solution
video by MIP4U 

Final Answer
\( 281994\sqrt{6} \)
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Surface Orientation
Before we discuss surface integrals over vector fields, we need to discuss surface orientation. Surface orientation is important because we need to know which direction the vector field is pointing, outside or inside, in order to determine the flux through the surface.
The two vectors that calculated above, \(\vec{r}_u\) and \(\vec{r}_v\) and tangent vectors to the surface. Using the cross product, we can calculate two possible normal vectors, \(\vec{N}_1 = \vec{r}_u \times \vec{r}_v\) and \(\vec{N}_2 = \vec{r}_v \times \vec{r}_u\). One vector points inward, the other points outward. We will use the first one, i.e. \(\vec{N}_1 = \vec{r}_u \times \vec{r}_v\) and divide by the length to get the unit vector \(\displaystyle{ \vec{N} = \frac{\vec{r}_u \times \vec{r}_v}{ \ \vec{r}_u \times \vec{r}_v \} }\) which is called the upward unit normal.
[Important Note: This may not always be the outward pointing normal but for our discussions we will work with surfaces where the outward pointing normal is this upward pointing normal. For your application, you will need to double check that you have the outward pointing normal.]
Surface Integrals of Vector Fields
In this video, Dr Chris Tisdell continues his discussion of surface integrals and talks about vector fields. Again, this is a great video to watch.
video by Dr Chris Tisdell 

Surface integrals over vector fields are often called flux integrals since we will often be calculating the flux through a closed surface. The flux of a vector field \(\vec{F}(x,y,z)=M(x,y,z)\hat{i}+N(x,y,z)\hat{j}+P(x,y,z)\hat{k}\) through a surface S with a unit normal vector \(\vec{N}\) is \( \iint\limits_S { \vec{F} \cdot \vec{N} ~ dS} \).
[ Note: In the above description, there are two N's. One is a function \(N(x,y,z)\) which is the jcomponent of the vector function. The other is \(\vec{N}\), a unit normal vector. They are distinct and unrelated and should not be confused. ]
When the surface is given in terms of \(z=g(x,y)\), we can calculate the unit normal vector as \(\displaystyle{ \frac{\nabla G}{\ \nabla G \} }\). Since \(dS = \ \nabla G \ ~ dA \) where \(G(x,y,z)=zg(x,y)\), the surface integral becomes \( \iint\limits_R { \vec{F} \cdot \nabla G ~ dA } \) where R is the projection of S in the xyplane.
Surface Integrals  Meaning and Applications
The meaning of the surface integral depends on what the function \(f(x,y,z)\) or \(\vec{F}(x,y,z)\) represents. Here is a video clip giving some applications.
video by Evans Lawrence 

In this final video, he gives more explanation of surface integrals and a couple of examples. He has a unique way of thinking about surface integrals and, as he says at the first of this video, surface integration is not easy. So it will help you to watch this video clip as well before going on to trying some on your own.
video by Evans Lawrence 

Okay, you are now ready for some practice problems calculating surface integrals using vector functions.
Then you will be ready for the three dimensional versions of Green's Theorem, Stokes' Theorem and the Divergence Theorem.
Practice  Surface Integrals, Vector Functions
Basic 

Compute the flux of \( \vec{F} = \langle x,y,z \rangle \) across the surface \( z = 4  x^2  y^2 \), \( z \geq 0 \) oriented up.
Problem Statement 

Compute the flux of \( \vec{F} = \langle x,y,z \rangle \) across the surface \( z = 4  x^2  y^2 \), \( z \geq 0 \) oriented up.
Final Answer 

\( 24\pi \)
Problem Statement
Compute the flux of \( \vec{F} = \langle x,y,z \rangle \) across the surface \( z = 4  x^2  y^2 \), \( z \geq 0 \) oriented up.
Solution
video by Evans Lawrence 

Final Answer
\( 24\pi \)
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Determine the flux of \( \vec{F} = \langle 0,1,2 \rangle \) across the surface \( z = 6  x  y \) in the first octant. Use a downward orientation.
Problem Statement 

Determine the flux of \( \vec{F} = \langle 0,1,2 \rangle \) across the surface \( z = 6  x  y \) in the first octant. Use a downward orientation.
Final Answer 

54
Problem Statement
Determine the flux of \( \vec{F} = \langle 0,1,2 \rangle \) across the surface \( z = 6  x  y \) in the first octant. Use a downward orientation.
Solution
video by MIP4U 

Final Answer
54
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Determine the surface area of the cylinder given by \( \vec{r} = \langle 3\cos(u), 3\sin(u), v\rangle \) for \( 0 \leq u \leq 2\pi \), \( 0 \leq v \leq 4 \).
Problem Statement 

Determine the surface area of the cylinder given by \( \vec{r} = \langle 3\cos(u), 3\sin(u), v\rangle \) for \( 0 \leq u \leq 2\pi \), \( 0 \leq v \leq 4 \).
Final Answer 

\( 24\pi \)
Problem Statement
Determine the surface area of the cylinder given by \( \vec{r} = \langle 3\cos(u), 3\sin(u), v\rangle \) for \( 0 \leq u \leq 2\pi \), \( 0 \leq v \leq 4 \).
Solution
video by MIP4U 

Final Answer
\( 24\pi \)
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Determine the surface area of the sphere given by \( \vec{r} = \langle 2\sin(u)\cos(v), 2\sin(u)\sin(v), 2\cos(u)\rangle \) for \( 0 \leq u \leq \pi \), \( 0 \leq v \leq 2\pi \).
Problem Statement 

Determine the surface area of the sphere given by \( \vec{r} = \langle 2\sin(u)\cos(v), 2\sin(u)\sin(v), 2\cos(u)\rangle \) for \( 0 \leq u \leq \pi \), \( 0 \leq v \leq 2\pi \).
Final Answer 

\( 16\pi \)
Problem Statement
Determine the surface area of the sphere given by \( \vec{r} = \langle 2\sin(u)\cos(v), 2\sin(u)\sin(v), 2\cos(u)\rangle \) for \( 0 \leq u \leq \pi \), \( 0 \leq v \leq 2\pi \).
Solution
video by MIP4U 

Final Answer
\( 16\pi \)
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Intermediate 

Calculate the flux of \( \vec{F} = z\hat{i} + yz\hat{j} + 2x\hat{k} \) across the upper hemisphere of the unit sphere oriented with outwardpointing normals.
Problem Statement 

Calculate the flux of \( \vec{F} = z\hat{i} + yz\hat{j} + 2x\hat{k} \) across the upper hemisphere of the unit sphere oriented with outwardpointing normals.
Final Answer 

\( \pi/4 \)
Problem Statement
Calculate the flux of \( \vec{F} = z\hat{i} + yz\hat{j} + 2x\hat{k} \) across the upper hemisphere of the unit sphere oriented with outwardpointing normals.
Solution
Equation 

Since the flux \(F\) is defined parametrically, we will use the equation \( \iint_{S}{ \vec{F} \cdot \vec{N} ~ dS } = \iint_{D}{ \vec{F} \cdot ( \vec{r}_u \times \vec{r}_v ) ~ dA } \) where \(\vec{N}\) is the unit normal vector. 

Surface 
The equation of a sphere is \(x^2 + y^2 + z^2 = R^2\) where \(R\) is the radius of the sphere. In our case, we have a unit sphere, so \(R=1\). It will be easier to convert to spherical coordinates and work with them, rather than rectangular coordinates. 
To go along with the notation in the tutorial, we will use the variables \(u\) and \(v\) instead of \(\theta\) and \(\phi\), where \(u=\theta\) and \(v=\phi\). This means the equation for the surface is \( \vec{r}(u, v) = \langle R\cos u \sin v, R\sin u \sin v, R\cos v \rangle \) where \(0 \leq u \lt 2\pi\) and \(0 \leq v \leq \pi/2\). 

Calculating The Partial Derivatives 
We need to determine \( \vec{r}_u \times \vec{r}_v \) where \( \vec{r}_u \) is the partial derivative of \( \vec{r} \) with respect to \(u\). Similarly for the second term. 
\( \vec{r}_u = \cos u \cos v \hat{i} + \) \( \cos v \sin u \hat{j}  \) \( \sin v \hat{k} \) 
\( \vec{r}_v = \sin v \sin u \hat{i} + \) \( \sin v \sin u \hat{j} + 0 \hat{k} \) 

Calculating The Cross Product 
The zero in the \(\hat{k}\) term of the second partial derivative helps to make the cross product not as complicated. 
\( \vec{r}_u \times \vec{r}_v = \) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \cos u \cos v & \cos v \sin u & \sin v \\ \sin u \sin v & \sin u \sin v & 0 \end{vmatrix} \) \( = \sin u \sin^2 v \hat{i} + \) \( \sin u \sin^2 v \hat{j} + \) \( ( \cos u \cos v \sin u \sin v + \cos v \sin u \sin^2 v ) \hat{k} \) 

Calculating The Dot Product 
Okay, so now we need to take the result from last line above and take the dot of that with the flux vector, i.e. we need to calculate \( \vec{F} \cdot ( \vec{r}_u \times \vec{r}_v ) \) 
The result, after a bit of simplifying is \( \cos v \sin^2 v \sin u + \cos v \sin^3 v \sin^2 u + 2\cos v \sin^2 v \cos^2 u \sin u + 2\cos v \sin ^2 v \cos u \sin^2 u \) 

Integrating 
Whew! Now we need to integrate that last equation twice, once with respect to \(u\) and again with resepect to \(v\). That seems like a lot of work. However, since the limits of integration in each case are all constants, we can easily choose the order of integration. In all four cases, we chose to integrate with respect to \(u\) first. This causes the integration of the first, third and fourth integrals to all go to zero without even needing to integrate in the other direction. The only integral that results in a nonzero value is the second one. We will go through that one and let you work the other three. 
The inside integral is \(\displaystyle{ \int_0^{2\pi}{ \sin^2 u ~ du } }\). Substituting \(\displaystyle{ \sin^2 u = \frac{1\cos(2u)}{2} }\) and integrating, gives us \( \pi \). 
The outside integral is then \(\displaystyle{ \int_0^{\pi/2}{ \pi \cos v \sin^3 v ~dv } }\) 
Pull a \(\sin v\) out of the cubed term, substitute \(\sin^2 u = 1 \cos^2 u\) and using integration by substitution yields \(\pi/4\) for the integral. 
Final Answer
\( \pi/4 \)
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