Cylindrical coordinates are useful for describing cylinders r = f(θ) z ⩾ 0 is the cylinder above the plane polar curve r = f(θ) r2 + z2 = a2 is the sphere of radius a centered at the origin r = mz m > 0 and z ⩾ 0 is the cone of slope m with cone point at the origin
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MATH 20550 Triple Integrals in cylindrical and spherical coordinates Fall 2016
1.Coordinates
1.1.Cylindrical coordinates.(r;;z)7!(x;y;z)
x=rcos y=rsin z=z Cylindrical coordinates are just polar coordinates in the plane andz. Useful formulas r=px 2+y2 tan=yx ;x6= 0;x= 0 =)=2 These are just the polar coordinate useful formulas. Cylindrical coordinates are useful for describing cylinders. r=f()z>0 is the cylinder above the plane polar curver=f(). r2+z2=a2
is the sphere of radiusacentered at the origin. r=mz m >0 andz>0 is the cone of slopemwith cone point at the origin.1.2.Spherical coordinates.(;;)7!(x;y;z)
x=cossin y=sinsin z=cosUseful formulas
=px2+y2+z2
sin=px 2+y2 tan=yx ;x6= 0;x= 0 =)=2 tan=px 2+y2z ;z6= 0;z= 0 =)=2 You can also change spherical coordinates into cylindrical coordinates. z=cos r=sin 1 2Standard graphs in spherical coordinates:
=a is the sphere of radiusacentered at the origin. =c is the cone of slope tan(c) with cone point at the origin. =c is the vertical plane over the liney= tan(c)x. cos=c is the planez=c. = 2dcos is the sphere of radiusjdjcentered at (0;0;d). = 2dcossin is the sphere of radiusjdjcentered at (d;0;0). = 2dsinsin is the sphere of radiusjdjcentered at (0;d;0).Indeed,
= 2acossin+ 2bsinsin+ 2ccos is the sphere of radiuspa2+b2+c2centered at (a;b;c).
2.Triple integrals
As usual, the goal is to evaluate some triple integral over some solid in space. If the solid isS, then ZZZ S f dV does not depend on any particular coordinate system (which is why I have not writtenf(x;y;z)). All you have to be able to do is to evaluatefat points inS(no matter how the points are described to you) and to compute the volume of the pieces into which you have partitionedS. Maybef is given to you in Cartesian coordinates,f(x;y;z), or maybe in terms of cylindrical coordinates, f(r;;z), or maybe in terms of spherical coordinates,f(;;). Given a formula in one coordinate system you can work out formulas forfin other coordinate systems but behind the scenes you are just evaluating a function,f, at a pointp2S. If you use a dierent coordinate system, the formula forflooks dierent but it is still the same function. If we agree on a pointp2Sand all go o and use our various formulas, we will all get the same number for the value offat that point. Aside:Remember that when we rst introduced the triple integral weestimatedthe triple integral just given a verbal description ofSand a table of values forf. We partitionedSso that in each piece we could choose one of the points from our table and then we wrote down the Riemann sum and that was our approximation. 32.1.Cylindrical coordinates.Suppose we have describedSin terms of cylindrical coordinates.
This means that we have a solidCin (r;;z) space and when we mapCinto space using cylindrical coordinates we getS. If we cutCup into little boxes we get little slices of pie in space soZZZ C f jrjdV=ZZZ S f dVTo use this formula usefully we will need to be able to evaluatefat points given to us in cylindrical
coordinates.2.2.Spherical coordinates.Suppose we have describedSin terms of spherical coordinates. This
means that we have a solid in (;;) space and when we map into space using spherical coordinates we getS. If we cut up into little boxes we get little pieces in space as described in the book ZZZ f2jsinjdV=ZZZ S f dV To use this formula usefully we will need to be able to evaluatefat points given to us in spherical coordinates. 42.3.Example.Suppose you want to integratex2over a ball of radiusacentered at the origin,ZZZ
S x2dV. In cylindrical coordinatesSis 06r6a, 0662,pa2r26z6pa
2r2. Hence ZZZ S x2dV=Z a 0Z 2 0Z pa 2r2 pa2r2r3cos2dz ddr
In spherical coordinatesSis 066a, 0662, 066. HenceZZZ S x2dV=Z a 0Z 2 0Z 04cos2sin3ddd
By now you should be able to see
ZZZ S x2dV=Z a aZ pa 2x2 pa 2x2Z pa 2x2y2 pa2x2y2x2dz dy dx
in Cartesian coordinates. I'm not overly excited about doing any of these integrals but the spherical coordinates one is the easiest. Since you are integrating over a box in (;;) space,Za 0Z 2 0Z 04cos2sin3ddd=
Za 0 4d Z2 0 cos2 d Z 0 sin3 d a 5523
=2a515 5
2.4.Example.Suppose you want the volume of the solid between the cones of slope 1 and slope12
and inside the cylinder over the circle of radius 3 centered at the origin in thexyplane,ZZZ S 1dV.In cylindrical coordinates 0662, 06r63 andr2
6z6rso
ZZZ S 1dV=Z 3 0Z 2 0Z r r2 rdz ddr In spherical coordinatesSis 0662,=466arctan(2). Thecoordinate starts at 0 and keeps going until it hits the cylinder. This happens whenr= 3 sosin= 3 or= 3csc.0663csc, HenceZZZ
S 1dV=Z 2 0Z arctan(2) =4Z 3csc 02sinddd
This time cylindrical looks easiest but not necessarily by much. Z3 0Z 2 0Z r r2 rdz ddr=Z 3 0Z 2 0 rzz=r z=r=2ddr=Z 3 0Z 2 0r 22ddr= 2Z 3 0r 22
dr= 2r36 3 0= 9
In spherical coordinatesZ2
0Z arctan(2) =4Z 3csc 02sinddd=Z
2 0Z arctan(2) =4 33sin=3csc =0dd= Z 2 0Z arctan(2) =49csc3sindd= 9Z 2 0Z arctan(2) =4csc2dd= 9 Z 2 0 cotarctan(2) =4d= 9Z 2
01=2 + 1d= 9
62.5.Example.Convert
Z 3 0Z p9y2 0Z p18x2y2 px2+y2(x2+y2+z2)dz dxdy
into a spherical coordinate iterated integral (from here , example 2.)Let us start by describing the solid. NoteR3
0Rp9y2
0dxdydescribes the quarter of the disk of
radius 3 in thexyplane in the rst quadrant. Above this quarter-disk thezcoordinate starts atz=px2+y2=rwhich is the cone of slope
1 and ends atz=p18x2y2which is the upper hemisphere of radiusp18 = 3
p2. Hence the solid is inside the cylinder parallel to thezaxis of radius 3, above the 45cone and below the sphere centered at the origin of radius 2p3. Hence in spherical coordinates one has 066=4, 066=2 and 0663p2. Hence Z 3 0Z p9y2 0Z p18x2y2 px2+y2(x2+y2+z2)dz dxdy=Z
4 0Z 2 0Z 3p2 04sin()ddd=
Z 4 0Z 2 0 553 p2 0 dd=4 2 2435
16p2 =
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