Triple Integrals in Cylindrical and Spherical Coordinates
Note: Remember that in polar coordinates dA = r dr d . EX 1 Find the volume of the solid bounded above by the sphere x2 + y2 + z2 = 9 below by the plane z
Review for Exam 3. Triple integral in spherical coordinates (Sect
2. ) hence A = 3π + 9. √. 3/2. <. Page 6. Double integrals in Cartesian coordinates. (Sect. 15.2). Example. Find the y-component of the centroid vector in
MATH 20550 Triple Integrals in cylindrical and spherical coordinates
into a spherical coordinate iterated integral. (from here example 2.) Let us start by describing the solid. Note ∫. 3. 0. ∫. √.
Triple Integrals in Cylindrical Coordinates Many applications involve
In particular there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example
15.8: Triple Integrals in Spherical Coordinates
We have: Page 6. Example. Let's get a better handle on things by graphing some basic functions given in spherical coordinates. Let c be a constant. Sketch the
April 8: Triple Integrals via Spherical and Cylindrical Coordinates
Apr 8 2020 Example 1. Let's begin as we did with polar coordinates. We want a. 3-dimensional analogue of integrating over a circle. So we integrate ...
Triple Integrals in Spherical Coordinates
For example the sphere with center the origin and radius c has the simple equation ρ = c (see Figure 2); this is the reason for the name “spherical”
Triple Integrals in Cylindrical and Spherical Coordinates
Triple Integrals in Cylindrical and Spherical Coordinates. 29/67. Page 30. How to Integrate in Spherical Coordinates - An Example. Example 5. Find the volume of
3.6 Integration with Cylindrical and Spherical Coordinates
In this section we describe
TRIPLE INTEGRALS IN SPHERICAL COORDINATES EXAMPLE A
TRIPLE INTEGRALS IN SPHERICAL COORDINATES. EXAMPLE A Find an equation in spherical coordinates for the hyperboloid of two sheets with equation . SOLUTION
Review for Exam 3. Triple integral in spherical coordinates (Sect
Line integrals in space. Example. Evaluate the line integral of the function f (xy
Page 1 Section 15.8: Triple Integrals in Spherical Coordinates
Spherical Coordinates: A Cartesian point (x y
15.8: Triple Integrals in Spherical Coordinates
We have: Page 6. Example. Let's get a better handle on things by graphing some basic functions given in spherical coordinates. Let c be a constant. Sketch the
MATH 20550 Triple Integrals in cylindrical and spherical coordinates
into a spherical coordinate iterated integral. (from here example 2.) Let us start by describing the solid. Note ?. 3. 0. ?. ?.
Integrals in cylindrical spherical coordinates (Sect. 15.7) Cylindrical
Notice the extra factor ?2 sin(?) on the right-hand side. Triple integral in spherical coordinates. Example. Find the volume of a sphere of radius R.
Triple Integrals in Cylindrical Coordinates Many applications involve
In particular there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example
Triple Integrals in Spherical Coordinates
For example the sphere with center the origin and radius c has the simple equation ? = c (see Figure 2); this is the reason for the name “spherical”
Triple Integrals in Cylindrical or Spherical Coordinates
xyz dV as an iterated integral in cylindrical coordinates. x y z. Solution. This is the same problem as #3 on the worksheet “Triple Integrals”
15.7 Triple Integrals in Cylindrical and Spherical Coordinates
Figure 15.44 Page 894. Example. Page 901
Section 9.7/12.8: Triple Integrals in Cylindrical and Spherical
from rectangular to spherical coordinates. Solution: ·. Example 7: Convert the equation ? ? sec2. =.
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 3quotesdbs_dbs14.pdfusesText_20[PDF] trivia questions pdf
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