MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use visit http ocw mit edu terms 18 02 Lecture 26 Tue Nov 13 2007 Spherical coordinates rho distance to origin phi angle down from z axis same as in cylindrical coordinates Diagram drawn in space and picture of 2D slice by vertical plane with z r coordinates Formulas to remember z cos r sin so x sin cos y sin sin x2 y 2 z 2 r2 z 2 The equation a de nes the sphere of radius a centered at 0 On the surface of the sphere is similar to latitude except it s 0 at the north pole 2 on the equator at the south pole is similar to longitude 4 is a cone asked using ash cards z r x2 y 2 2 is the xy plane Volume element dV 2 sin d d d To understand this formula rst study surface area on sphere of radius a picture shown of a rectangle corresponding to with sides portion of circle of radius a of length a and portion of circle of radius r a sin of length r a sin So S a2 sin which gives the surface element dS a2 sin d d The volume element follows for a small box V S so dV d dS 2 sin d d d Example recall the complicated example at end of Friday s lecture region sliced by a plane inside unit sphere After rotating coordinate system the question becomes volume of the portion of unit sphere above the plane z 1 2 picture drawn This can be set up in cylindrical left as exercise or spherical coordinates For xed we are slicing our region by rays straight out of the origin ranges from its value on the plane z 1 2 to its value on the sphere 1 Spherical coordinate equation of the plane z cos 1 2 so sec 2 The volume is 2 4 1 2 sin d d d 0 0 1 2 sec Bound for explained by looking at a slice by vertical plane constant the edge of the region is at z r 12 Evaluation not done Final answer 2 5 3 6 2 Application to gravitation Gravitational force exerted on mass m at origin by a mass M at x y z picture shown G M m x y z G M m is given by F dir F i e F x y z G gravitational 3 2 constant If instead of a point mass we have a solid with density then we must integrate contributions to gravitational attraction from small pieces M V So Gm x y z z dV i e z component is F Gm dV F z 3 3 R R If we can set up to use symmetry then Fz can be computed nicely using spherical coordinates General setup place the mass m at the origin so integrand is as above and place the solid so that the z axis is an axis of symmetry Then F 0 0 Fz by symmetry and we have only one 1 2 component to compute Then z cos 2 Fz Gm dV Gm sin d d d Gm cos sin d d d 3 3 R R R Example Newton s theorem the gravitational attraction of a spherical planet with uniform density is the same as that of the equivalent point mass at its center Setup the sphere has radius a and is centered on the positive z axis tangent to xy plane at the origin the test mass is m at the origin Then 2 2 2a cos 4 GM m z Fz Gm dV Gm cos sin d d d Gm a 3 3 a2 R 0 0 0 where M mass of the planet 43 a3 The bounds for and need to be explained carefully by drawing a diagram of a vertical slice with z and r coordinate axes and the inscribed right triangle with vertices the two poles of the sphere a point on its surface the hypothenuse is the diameter 2a and we get 2a cos for the spherical coordinate equation of the sphere 18 02 Lecture 27 Thu Nov 15 2007 Handouts PS10 solutions PS11 Vector elds in space At every point in space F P Qj Rk where P Q R are functions of x y z Examples force elds gravitational force F c x y z 3 electric eld E magnetic eld B velocity elds uid ow v v x y z gradient elds e g temperature and pressure gradients Flux Recall in 2D ux of a vector eld F across a curve C C F n ds In 3D ux of a vector eld is a double integral ux through a surface not a curve F vector eld S surface n unit normal vector Flux F n dS is often easier to compute than n and dS n dS We ll see that dS Notation dS Remark there are 2 choices for n choose which way is counted positively Geometric interpretation of ux As in 2D if F velocity of a uid ow then ux ow per unit time across S Cut S into small pieces then over each small piece what passes through S in unit time is the contents of a parallelepiped with base S and third side given by F Volume of box base height F n S Examples 1 F x yj zk through sphere of radius a centered at 0 n a1 x y z other choice a1 x y z traditionally choose n pointing out F n x y z n a1 x2 y 2 z 2 a so S F n dS S a dS a 4 a2 3 n z 2 zk H 2 Same sphere H a 2 2 2 z 4 a cos2 2 dS H dS a sin d d 2 a3 cos2 sin d a3 3 a a S S 0 0 0 Setup Sometimes we have an easy geometric argument but in general we must compute the surface integral The setup requires the use of two parameters to describe the surface and F n dS must be expressed in terms of them How to do this depends on the type of surface For now formulas to remember 0 plane z a parallel to xy plane n k dS dx dy similarly for planes xz or yz plane 1 sphere of radius a centered at origin use substitute a for evaluation n a1 x y z dS a2 sin d d 2 cylinder of radius a centered on z axis use z substitute r a for evaluation n is radially out in horizontal directions away from z axis i e n a1 x y 0 and dS a dz d explained by drawing a picture of a rectangular piece of cylinder S z a 3 graph z f x y use x y substitute z f x y We ll see on Friday that n and dS separately are complicated but n dS fx …
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