18 02 Multivariable Calculus Spring 2009 Lecture 31 Surface Area and flux across cylinders and spheres April 24 Reading Material From Simmons 21 4 From Lecture Notes V8 V9 Last time Surface Area Vector fields in 3D Surface Integrals and flux Today Surface Area and flux across cylinders and spheres In the last class we learned that the flux across an oriented surface S is ZZ flux of F across S F n dS S so you could also write Shorthand We often use the notation n dS dS ZZ F dS flux of F across S S This is a typical surface integral We already know how to compute one of then namely the area of a surface ZZ Area S 1dS S so it will not be difficult to generalize this formula 2 Evaluating flux Integrals Assume S is a surface given by the graph of a function f x y Assume S is oriented by using the upward normal To convert the flux integral ZZ F dS S into a double iterated integral we use 3 2 above and we observe that n dS dS N dA f xi f y j k dA N N 1 and as a consequence we simply have ZZ ZZ F dS F x y f x y f xi f y j k dA S R where R is the shadow cast by S on the xy plane Exercise 1 Consider the vector field F z k and the surface S given by the graph of z x2 y 2 above the region R x y 1 x y 1 oriented upward Compute the flux of F across S Solution Exercise 2 Assume that now F xj and S is given by the graph of z 1 x2 y 2 for x y z 0 and S is oriented outward Again compute the flux of F across S Solulition 2 Last time we gave formulas to compute the area of surfaces and flux across a surface given as graphs of functions of type z f x y In this section we consider instead a piece of a cylinder and a piece of a sphere 3 Surface area and flux cylinders We first recall that the cylindrical coordinates are obtained by using polar coordinates in the xy plane and the z coordinate along the z axis to summarize rectangular x y z cylindrical polar z r z Conversion Like polar r z x y z x r cos y r sin z z and the infinitesimal element of volume is dV r d dr dz 3 1 It is clear that if we want to describe the piece of cylinder C of radius a and hight h between the angles 1 and 2 we just need to use the coordinates above where we FIX r a To measure its area we have to compute ZZ Area 1dS C and from 3 1 it is not hard to deduce that dS a d dz since r a and there is no variation in r As a consequence Zh Z 2 ZZ Area a d dz a 2 1 h 1dS C 0 1 3 3 2 as expected Now assume the cylinder is oriented outward then it is clear that at the point P a cos a sin z the outward normal vector n cos i sin j since at each hight we have the picture Given a vector filed F M i M j P k the flux through the oriented cylinder D of F is ZZ Flux C Zh Z 2 M a cos a sin z cos N a cos a sin z sin a d dz F n dS 0 1 Note For simplicity here I am assuming that the cylinder is touching the xy plane in other words 0 z h But it is also possible to have a more general situation when h1 z h2 for some arbitrary h1 and h2 Exercise 3 Consider the vector field F xi z k and the cylinder C where 0 0 z h and a 1 We orient it outward Compute the flux of F across C 4 Solution 4 Surface area and flux spheres where r z x y z r sin z cos x r cos sin cos y r sin sin sin z z cos We also recall that dV 2 sin d d d 4 1 Now consider for example the surface S given by the piece of sphere centered at the origin of radius a with longitude between 1 and 2 and latitude between 1 and 2 5 To compute the area of its surface we deduce from 4 1 that dS a2 sin d d 4 2 since there is no variation in and we write Z 2 Z 2 ZZ Area 1 dS S a2 sin d d 1 1 Suppose now we orient S by outward normal n To find this n we first reason in rectangular coordinates Since the sphere is described by x2 y 2 z 2 a2 0 it is easy for example by taking the gradient that a normal outward vector is xi y j z k N The size of this vector is N p x2 y 2 z 2 a on the sphere hence n 1 xi y j z k a We can then write n by using spherical coordinates as n sin cos i sin sin j cos k 4 3 Now assume a vector field F M i M j P k is also given and we want to compute its flux across S We have from 4 2 and 4 3 that ZZ F n dS Flux S Z 2 Z 2 M a sin cos a sin sin a cos sin cos N sin sin P cos a2 sin d d 1 1 where we use for short instead of rewriting two more times the same coordinates that appear in the argument of M Exercise 4 Consider the piece S of the sphere centered at the origin radius 2 with 0 2 and 0 2 oriented inward Compute the flux of the vector field F xi across the surface S 6 Solution Handy Fact In general given an oriented surface S with normal vector n and a vector field F if F n C for some constant C then ZZ Flux F n dS C Area S S 7
View Full Document
Unlocking...