Auld Lecture 2

7/25/2019 Auld Lecture 2

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Spherical and Cylindrical Coordinates

Eric Auld

January 14, 2016


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Bellwork

Sketch the region of integration of the following integral, andthen reverse the order of integration to get an equivalentintegral: y=1

y=0

x=yx=y

dx dy

Suppose we have a function f(x, y) which we want tointegrate over the triangle with vertices at (1, 0), (1, 0), and

(0, 1). Set up the integral.


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Solution to 1Draw the graph ofx=

y.

y=1

y=0

x=

y

x=ydx dy

Draw the graph ofx=y.

We are proceeding first horizontally, and then vertically. If we wantto proceed first vertically and then horizontally, we need y between

x2 and x, and x between zero and one. So the integral in theopposite order is

10

xx2

dy dx.


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Solution to 2

We need to represent the boundary as graphs. The upper-left partof the triangle is the graph of the function y=x+ 1, or

x=y 1. Similarly, the upper-right part is the graph ofy= 1 x, orx= 1 y. Therefore, in the dx dyway, theintegral is

y=1

y=0

x=1y

x=y1

f(x, y) dx dy.

In the latter case, the integral is

x=0

x=1

y=x+1

y=0f(x, y) dy dx+

10

1x0

f(x, y) dy dx.


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Polar Coordinates

Express as a graph in polar coordinates the circle centered at(0, 1).

While this figure is nota graph of a function in xy coordinates, we

will see that it is the graph of a function in polar coordinates.First, lets recall that the equation describing this figure in xycoordinates is x2 + (y 1)2 = 1.

Method 1: Plug in for x and y, solve and get r= 2 cos .

Method 2: Geometrically


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Exercise 16.4.14

Sketch the region of integration and evaluate by changing to polar

coordinates: 21

2xx20

1x2 +y2

dy dx

We need to draw the graphs of y= 0 and y=

2x x2. Towork with y= 2x x2, we try to square both sides, givingy2 = 2x x2. When we have a pesky linear term we want to getrid of, the solution is to complete the square.2x x2 = (x2 2x) = (x 1)2 + C= (x 1)2 + 1 after wesolve for C. So y2 =

(x

1)2 + 1, or y2 + (x

1)2 = 1, and that

we do know how to draw. It is a circle of radius 1 centered at(1, 0).

How can we represent this in polar coordinates?


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Exercise 16.4.14 Continued

Our shape is the right half of a disk of radius one centered at (1, 0).We first need to figure out what functions have the graphs of

The circle of radius one centered at the (1, 0). (We just didthis one

The straight line x= 1. (Notice we know the function thathas that graph in rectangular coordinates, namely x= 1, butwe dont yet know in polar what it is.

Again, the same two methods work: plugging and chugging, or

working geometrically.


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Exercise 16.4.14 Continued

The region is given in polar coordinates by

0 /4; sec r 2cos .

So we would write the integral as

/40

2cos sec

1

r (r dr d)

=

/40

2cos sec

dr d,

which is not hard to evaluate (if you remember how to integratesecant).


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Questions

1. Remind yourself of the material we covered last time: whatare the points constant distance away from the origin, whatare the points whose ycoordinate is the same as theirdistance from the yaxis, etc.

2. Characterize the paraboloid opening along the positive x axisin wordsIt is the points in three dimensional space such thatthe distance...

3. Remind yourself from last time what happens when one

variable does not appear in the description of a figure.


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Exercise 16.3.23Evaluate

Wxz dV, whereW is the domain bounded by the

elliptic cylinder x2

4 + y

2

9 = 1 and the sphere x2 +y2 +z2 = 16, in

the first octant. We need to figureout which direction we want to integrate in first. In whichdirection is the region bounded by two simple graphs? If we were

to try to solve for the variables in the expressions x2

4 + y2

9 = 1 andx2 + y2 + z2 = 16, we would take square roots, and we would haveto choose either the plus or the minus side, if we wanted to end upwith a graph. But because were in the first octant, our x, y and z

values are going to be positive. That means that we dont have toworr we can ust take the ositive branch in ever case.


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Exercise 16.3.23, part 2

How can we parameterize this region? A: 0 y 3

1 (x/2)2,0 x 2. So the integral should be

2

0 3

1(x/2)2

0

16x2y2

0

xz dz dy dx


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20

3

1

(x/2)2

0

16

x2

y2

0xz dz dy dx

=

20

31(x/2)20

1

2xz2

16x2y2

0

dy dx

=

20

31(x/2)20

1

2x(16 x2 y2)dy dx

= 20

31(x/2)20

8x12

x3 12

xy2dy dx

=

20

8xy 1

2x3y 1

6xy3

31(x/2)20

dx


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20

8xy 1

2x3y 1

6xy3

31(x/2)20

dx

= 2

0 24x

1 (x/2)2

3

2 x

31 (x/2)

2

276

x(1 (x/2)2)

1 (x/2)2 dx,

which Im not going to solve for you. There are two integrals in

there: one is of the form x1 x2 (up to constant factors), andthat one you can solve by substitution. The other one is of theform x3

1 x2. How would you integrate that? A:Parts: take

dv=x

1 x2 and u=x2.


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Cylindrical CoordinatesThere are two popular ways to describe coordinates inthree-dimensional space besides just stating the x, y, and z

coordinates. They are called cylindrical coordinates and sphericalcoordinates. Well talk about cylindrical coordinates first.

One popular way is to pick a particular axis (could be the x, y, orzaxis...suppose we choose the zaxis), and tell

1. The distance from the zaxis (call it r).2. The z coordinate.

3. The direction from the z axis.

This is called cylindrical coordinates. The name comes from the

fact that the points with common distance from an axis form acylinder, and that shape is particularly natural in these coordinates(just the points r=c). Again we have three coordinates.

Essentially what were doing with cylindrical coordinates is pickinga plane z=a and then using polar coordinates in that plane.


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Cylindrical Coordinates 2To restate, now were using three coordinates, r, the distance from the z axis,

, the direction from the zaxis, can be any number, butmany numbers refer to the same direction (those that are 2apart from each other). If you want to have unique numbersfor each direction, you can use thetas in [0, 2).

z, the usual z coordinate.

Now we need to figure out how to talk about functions of (r, z, )instead of functions of (x, y, z). But the nice thing about this isthat some shapes that formerly were not graphs are now graphs!

Example: The cylinder (say, the one hugging the zaxis) cannot

be a single graph in (x, y, z) coordinates, because it does not passany of the line tests (for example, it cannot be a graph z=f(x, y)because the cylinder does not pass the vertical line test. Anyvertical line that meets the cylinder would meet it in more that onepoint, which is not allowed. (In fact, it would meet it in infinitelymany points.))


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Cylindrical Coordinates 3

But the cylinder isa graph of a single function in cylindrical

coordinates! It is the graph of the simple function r=c. Noticehere that and zdo not appear at all.


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Figures in Which One Variable Does Not AppearWe describe regions (using shorthand) by saying for instance thesurface z=

x2 +y2.

Since we have the points in three-dimensional space such that...hiding before our description of the figure, if one variable does notappear, that means there are no constraints on that variable.That is, the value of that variable is not at all restricted.

Geometrically, ifydoes not appear in the figure, for instance, wecan move freely back and forth in the ydirection, and we will stayon the figure if we were already in it.

Example: What figure in 3D is described by x2 +y2 =c?

A: Since we see x2 +y2 in there, we should think of that as thesquare of the distance from the zaxis. The square of the distanceis a constant exactly when the distance itself is a constant(distance cant be negative, so there is only one square root), sowhat were really saying is that these are all the points of aparticular distance away from the zaxis. That distance isc.

C C


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Cylindrical Coordinates 3

But the cylinder isa graph of a single function in cylindricalcoordinates! It is the graph of the simple function r=c. Noticehere that and z do not appear at all. That means thatchanging and zdoes not change whether or not were on thefigure. In other words, the and z coordinates do not matter for

whether were on the figure or notonly the rcoordinate matters,and all that matters is that r=c.

What if we wanted our cylinder to hug the yaxis instead of the zaxis, and we wanted to express it as a single graph? A:We

would have to reframe our coordinates. We would choose r to bethe distance from the y axis, to be the direction from the y axis,and our third coordinate would be y.

S h i l C di


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Spherical Coordinates

Another popular way to describe points in three-dimensional space

is to tell

1. The distance from the origin (call it by the Greek letter ,rho)

2. The direction from the origin (described by Greek letters ,

phi, pronounced fye or fee, and and , theta).This is called spherical coordinates. The name comes from the factthat the sphere is the points with a common distance from theorigin, and this shape is particularly natural to describe in thesecoordinates (=c).

To describe the direction we need two coordinates, so we havehave three coordinates, just like before.

S h i l C di 2


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Spherical Coordinates 2

The tricky thing about spherical coordinates is understanding howthe two parameters and describe the direction from the

originand why has 360 degrees of possibilities while only has180 degrees of possibilities.

Notice that describing a direction from the center is exactly the

same as fixing a sphere centered at the origin and describing pointson that sphere. (Convince yourself that for every direction fromthe origin, there is a unique point on that sphere that you wouldhit if you traveled that direction. Conversely, for every point onthat sphere, there is a unique direction youd have to travel in a

straight line to hit that point. This gives us a useful analogy tothink of how to describe directions from the origin: on Earth wedescribe our position on a sphere by two parameters: longitude andlatitude! And notice that longitude has 360 degrees of possibilities,while latitude only has 180 degrees of possibilities.


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What if I wanted to describe a paraboloid, opening along thepositiveyaxis, in cylindrical coordinates? A:The paraboloid isthe points where the ycoordinate is the square of the distance

from they

axis. It benefits us to choose cylindrical coordinateswith respect to the yaxis, since this function involves distancefrom the yaxis. If we choose coordinates r, , y, then

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