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**Surface Area and Surface Integrals**

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Surface Area Given some surface in 3 space, we want to calculate its surface area Just as before, a double integral can be used to calculate the area of a surface We are going to look at how to calculate the surface area of a parameterized surface over a given region

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**Given the vector parameterization the surface area is given by**

Let’s take a look at where this comes from Example Find the surface area of a cone with a height of 1 The parameterization is Let’s check it out in maple

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Alternative Notation If we want to find the surface area of a function, z = f(x,y), than we can simplify the cross product Then and

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Alternative Notation If we want to find the surface area of a function, z = f(x,y), than we can use the following Example Find the surface area of the plane z = 6 – 3x – 2y that lies in the first octant

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**We can calculate the surface area over any given region Example**

Find the surface area of the function z = xy between the two cylinders

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Surface Integrals A surface integral involves integrating a function over some surface in 3 space We have calculated integrals of functions over regions in the xy plane and over 3 dimensional figures, now we want to integrate over a 2 dimensional surface in 3 space Thus if the function represents a density, the surface integral would calculate the total mass of the 2D plate that has the shape of the surface

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Surface Integrals To calculate a surface integral of g over the surface D if the surface is defined parametrically we have Example Calculate the surface integral of f(x,y) = xy over the cone of radius 1 in the first octant from the previous example

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Surface Integrals To calculate a surface integral of g over the surface D if the surface is given by z = f(x,y) we can use Example Find the surface integral of the function g(x,y,z) = xyz over the plane z = 6 – 3x – 2y that lies in the first octant

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**Surface Integrals of Vector Fields**

Recall that a line integral of a vector field could be interpreted as work done by the force field on a particle moving along the path If the vector field represents the flow of a fluid, then the surface integral of will represent the amount of fluid flowing through the surface (per unit time) In this case the amount of fluid flowing through the per unit time is called the flux Surface integrals of a vector field are sometimes referred to as flux integrals

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**Surface Integrals of Vector Fields**

The term flux comes from physics It is used to denote the rate of transfer of: Fluid liquid flow density Particles Electromagnetism Energy across a surface Total charge of a surface

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**Surface Integrals of Vector Fields**

Imagine water flowing through a surface If the flow of water is perpendicular to the surface a lot of water will flow through and the flux will be large If the flow of water is parallel to the surface then no water will flow through the surface and the flux will be zero In order to calculate the flux we must add up the component of that is perpendicular to the surface

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**Surface Integrals of Vector Fields**

Let represent a unit normal vector to the surface Than in order to find the component of that is perpendicular we can use our dot product This is 0 if and are perpendicular Positive if and are in the same direction Negative if and are in opposite directions Given some fluid flow , integrating will determine the total flux of fluid through a surface It will be positive if it is in the same direction as Negative if it is in the opposite direction of

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**Surface Integrals of Vector Fields**

Now we must sum over our surface so we will combine our dot product with our formula for a surface integral from before and we get the following This can be simplified!

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**Surface Integrals of Vector Fields**

The formula for a unit normal vector given our surface parameterization is Inserting that into our surface integral we get

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**Surface Integrals of Vector Fields**

We can cancel scalars to get Example The surface will be the parabaloid , 0 ≤ z ≤ 1 with the vector field Should our integral be positive or negative? How can we tell?

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**Surface Integrals of Vector Fields**

In order for a surface to have an orientation the surface must have two sides Thus every point will have two normal vectors, The set we choose determines the orientation which is described as the positive orientation You should be able to choose a normal vector in a way so that if it varies in a continuous way over the surface, when you return to the initial position it still points in the same direction

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**The Möbius band is not orientable**

No matter where you start to construct a continuous unit normal field, moving the vector continuously around the surface will return it to the starting point with a direction opposite to the one it had when it started.

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**Surface Integrals of Vector Fields**

As mentioned before, a surface integral over a vector field is positive if the normal of the surface and flow are in the same direction, negative if they are in opposite directions and 0 if they are perpendicular How do we know which normal to use for a surface? A surface is closed if it is the boundary of some solid region For example the surface of a sphere is closed A closed surface has a positive orientation if we choose the set of normal vectors that point outward from the region A closed surface has a negative oritenation if we choose the set of normal vectors that point inward toward the region This convention is only used for closed surfaces The surface in our previous example was not closed so this does not apply

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**Surface Integrals of Vector Fields**

In order to calculate our surface integral we use Since is a normal vector to the surface we can rewrite the integral as Now if our surface is given by a function z = f(x,y) then where f(x,y,z) = f(x,y) - z and our integral becomes Let’s try our previous example again with this method

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**Surface Integrals of Vector Fields**

Calculate the flux of of the surface S which is a hemisphere given by the following In this case we have a closed bounded region so our surface has a positive orientation that is pointing outwards Should we expect our integral to be positive or negative? In order to calculate this integral we will have to break S into 2 separate regions

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**Relationship between Surface Integrals and Line Integrals**

To calculate a line integral we use which summed up the components of the vector field that were tangent to the path given by To calculate a surface integral we use which sums up the components of the vector field that are in the normal direction given by

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