Report

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 • We are going to look at how to calculate the surface area of a parameterized surface over a given region • Given the vector parameterization r (u, v) x(u, v), y(u, v), z(u, v) the surface area is given by R ru rv dudv 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 r cos , r sin , r – 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 r x, y, z x, y, f ( x, y) rx 1,0, f x ( x, y ) and ry 0,1, f y ( x, y ) rx ry 1,0, f x 0,1, f y f x , f y ,1 ( f x ) ( f y ) 1 2 2 Alternative Notation • If we want to find the surface area of a function, z = f(x,y), than we can use the following R ( f x ) ( f y ) 1 dxdy 2 2 • 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 x y 1 2 2 x y 4 2 2 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 Surface Integrals • To calculate a surface integral of g over the surface D if the surface is defined parametrically we have R g ( r (u, v )) ru rv dudv • 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 R g ( x, y, f ( x, y )) ( f x ) ( f y ) 1 dxdy 2 2 • 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 F could be interpreted as work done by the force field on a particle moving along the path • If the vector field F represents the flow of a fluid, then the surface integral of F 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 F that is perpendicular to the surface Surface Integrals of Vector Fields • Let n represent a unit normal vector to the surface • Than in order to find the component of F that is perpendicular we can use our dot product F n – This is 0 if F and n are perpendicular – Positive if F and n are in the same direction – Negative if F and n are in opposite directions • Given some fluid flow F, integrating F n will determine the total flux of fluid through a surface – It will be positive if it is in the same direction as n – Negative if it is in the opposite direction of n 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 R g ( r (u, v )) ru rv dA and we get the following F n r r R • This can be simplified! u v dA Surface Integrals of Vector Fields • The formula for a unit normal vector given our surface parameterization r is ru rv n ru rv • Inserting that into our surface integral F n r r R we get u v dA ru rv F r r dA u v R r r u v Surface Integrals of Vector Fields • We can cancel scalars ru rv F r r dA u v R r r u v to get • Example F r r dA R u v 2 2 z x y – The surface will be the parabaloid ,0≤z ≤ 1 with the vector field F x, y,0 – 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, n1 and n2 n1 • 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 F r r dA u R v • Since ru rv is a normal vector to the surface we can rewrite the integral as F n dA R • Now if our surface is given by a function z = f(x,y) then n f ( x, y, z) where f(x,y,z) = f(x,y) - z and our integral becomes F f ( x, y, z) dA R • Let’s try our previous example again with this method Surface Integrals of Vector Fields • Calculate the flux of F x, y, z of the surface S which is a hemisphere given by the following 2 x y z 1 2 2 2 x2 y2 1 • 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 F (r (t )) dr where dr x '(t )dt, y '(t )dt C which summed up the components of the vector field that were tangent to the path given by dr • To calculate a surface integral we use R F ru rv dA which sums up the components of the vector field that are in the normal direction given by ru rv