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Surface Area and Surface Integrals. 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

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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**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**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**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**We can calculate the surface area over any given region**• Example • Find the surface area of the function z = xy between the two cylinders**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 xyplane 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**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**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**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**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**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**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**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!**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**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?**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**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.**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**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**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**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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