GRAPHING POLAR EQUATIONS

CLASSROOM SUPPLEMENT – APPENDIX H.1 1

GRAPHING POLAR EQUATIONS Example 1: Graph 2cos( )r , using the following directions: (i) By plotting points, using a table of values for ,r .

(ii) By using technology:

DIRECTIONS FOR POLAR GRAPHING USING Maple, . In Maple, type in the polar equation r = 2cos( ). Then hit ENTER. Right click on the blue output equation (Ctrl & click on a MAC) then go to PLOTS then PLOT BUILDER. First change the -parameter’s range to 0 to pi. Then click on OPTIONS then change the coordinate system from ‘Cartesian’ to ‘Polar with Polar Axes.’ Finally, hit Plot.


DIRECTIONS FOR POLAR GRAPHING USING A TI-83

Press the MODE key and arrow down to the fourth row where Func (function) is highlighted. Right arrow over to Pol (polar) and press ENTER. Press the Y= key and in \r1= type 2cos( ). Press WINDOW and type in 0 for min , for max ; use 3 3x and 2 2y . Press GRAPH.

DIRECTIONS FOR POLAR GRAPHING USING A TI-89

Press the MODE key and arrow to the right and then down to #3: POLAR. Press ENTER. Press the Y= key and in r1= type 2cos( ). Press WINDOW and type in 0 for min , for max ; use 3 3x and 2 2y . Press GRAPH.


POLAR GRAPHS AS PARAMETRIC CURVES

THE SLOPE dydx

OF A POLAR CURVE

Recall that the parametric representation of a curve uses two equations ( ), ( )x x t y y t where t is called a parameter. There are other parameters such as which we will use together with the equations x r cos() and y r sin() , to express a polar curve ( )r f , parametrically, as

( ) ( ) cos( ) and ( ) ( )sin( )x f y f . Example 1: (i) Express the polar equation, 2cos( )r parametrically. (ii) Use Maple to graph the set of parametric equations. You should get the

same circle found previously by graphing the polar equation 2cos( )r . Type in [2(cos( ))2,2sin( ) cos( )] . Then hit ENTER. Right click on the blue output equation (Ctrl & click on a MAC) then go to PLOT then PLOT BUILDER. First change the -parameter’s range to 0 to pi. Select 2-D Parametric Plot then hit the Plot button.

The purpose for writing a polar equation parametrically is to be able to find the slope of a polar

graph. Recall that

dydy d

dxdxd

if ( ), ( )x y is given (1)

so we will write ( )r r as ( ) cos( )x r ( )sin( )y r then using (1) and the product rule for differentiation, we have

( ) cos( ) ( )sin( )( )sin( ) ( ) cos( )

dy r rdx r r


Example 2: Find the slope dydx

of 2cos( ) at 3

r . Draw a tangent line at 3 on

the graph you sketched in Example 1.

What is drd

? What information does drd

= 0 give about the graph of 2cos( )r ?

Class Practice Exercises:

a) Sketch the graph of the polar equation 3 2cos( )r by using Maple or your graphing

calculator.

b) Find the slope dydx

of the polar curve 3 2cos( )r at 2 . Sketch the tangent line to

the curve at the point where 2 .

c) Find drd

and then solve drd

= 0. Check that the extrema of 3 2cos( )r

occur at the values of for which drd

= 0.


POLAR COORDINATES AND INTEGRATION AREA Example 1: (i) Sketch the graph of sin(3 )r by using Maple.

(ii) Find the area enclosed by one leaf (petal). Using the formula

21 ( )2

b

aA r d , the required integral is

/ 3 2

0

1 sin (3 )2

d

or, using

symmetry, / 6 2

0sin (3 )d

. Use Maple to do the integration. You can use

the integration template in the Expression palette. Maple gives you an

exact answer of 1

12 . To get an approximation, right click on the blue

output (Ctrl & click on a MAC) then choose Approximation and the number of digits you’d like Maple to approximate it to.

Directions for using a graphing calculator to integrate.

TI-83: MATH arrow down to #9 fnInt( and press ENTER (or just press MATH and type 9) Now type .5(sin(3X))^2,X,0, / 3 ) press ENTER. Note: if you are in Polar Mode, a will come up instead of an X when you hit the X,T, ,n key. Also, the commas are important. The comma key is next to the 2x key. The answer is shown as .2617993878

TI-89: On the entry line type 2nd 7 (this is the sign) Then type .5(sin(3x))^2,x,0, / 3 ) ENTER . The answer is shown as .261799, if in MODE you have Display Digits as FLOAT 6.


TANGENT LINES AT THE POLE Sometimes it is necessary (or helpful) to find tangent lines at the pole: If

( ) 0 and ( ) 0f f , then the line is tangent at the pole to the graph of ( )r f . Example 2: Find the area of the region lying inside the inner loop of the limaçon

1 2sin( )r . Use the following steps: (i) Graph 1 2sin( )r using Maple.

(ii) Set up the integration for the area. Use exact values for the limits of

integration.

(iii) Use Maple to carry out the integration.

CLASSROOM SUPPLEMENT – SECTIONS 9.2 – 9.4 7

VECTORS Using Maple It’s very easy to create a vector/matrix in Maple. Go to the Matrix palette and choose the number of rows and columns you want. Then click Insert Matrix (or Insert Vector if either the # of rows or columns is 1). Let’s as an example, enter the row vector [1,0,-1]. Hit the tab button between entries in the matrix. There may be times when you want to give a vector a letter name to refer to it or to use it in formulas. Don’t forget that to name an object in Maple you type the name followed by a colon followed by the equal sign. For example, typing a:=[1,0,-1] into Maple assigns the name ‘a’ to our row vector. Maple will compute the length of a vector (i.e. the Euclidean-norm), the dot and cross products of two vectors. Maple will also compute the determinant of a square matrix. (The TI-82/83 calculators can only compute the determinant of a square matrix. These calculators will not compute dot and cross products.) Let us find the length of a vector by using Maple’s built in function, Euclidean-norm. Right click on the blue output vector (Ctrl & click on a MAC) then go to NORM then EUCLIDEAN. Maple returns the answer 2 . To compute the dot product of the vectors [-1,1,3] and [2,2,-1/2], follow these steps: Enter the vectors in the way mentioned above, also giving them names of u and v. Then to get

the dot product of u and v type u.v and hit ENTER . Maple returns the answer

32

. Please

note that the dot between u and v in the syntax is a period, NOT a multiplication sign. To compute the angle between these two vectors follow these steps: Load Maple’s Vector Calculus package by clicking on Tools then Load Packages then Vector Calculus. Now enter the

expression, arccosu.v

u v

. Maple will return the answer 1.72891 (which is the angle in

radians). Please note that the operation in the numerator is the dot product and the operation in the denominator is multiplication between the Euclidean norms of u and v. The double vertical bars which compute the Euclidean-norm (length) of the vector named inside are found in the third to last row of the operators palette. The double bars are this operator ONLY when the Vector Calculus package is invoked. To compute the cross product of the vectors [-1,1,3] and [2,2,-1/2], follow these steps: Maple still recognizes these vectors as the vectors u and v so type u v and hit ENTER . Maple

will return the answer 132

,112

,4

. Please note that the cross product operator, , is not the

letter ‘x.’ To get this operator in Maple, click on the Operators palette. The cross product operator is located in the second row.


To compute the determinant of a 3x3 matrix enter the matrix using the matrix palette. For

example, enter the matrix 2 6 20 4 22 2 4

. To get its determinant right click on the blue output

matrix (Ctrl & click on a MAC) then go to Standard Operations then Determinant. Maple will return the answer -16. To enter matrices on the TI-83: Press the MATRIX key. Right arrow over to EDIT. Choose the letter you want, for instance, [A] by pressing ENTER. You can then choose the dimensions, for instance, 3 3, ENTER. Type in each entry and press ENTER after each one. When you have the matrix completed, press 2nd QUIT to return to the homescreen. Press the MATRX key

and with [A] highlighted press ENTER twice. You should see [[ ][ ][ ]]

2 6 20 4 22 2 4

on the screen. Then

press MATRX right arrow over to MATH press ENTER so that det( appears on the homescreen. Press 2nd ANS ENTER and – 16 appears on the screen. The TI-82/83 will not compute a dot or cross product. Matrices on the TI-89: There are two ways to enter a matrix on the TI-89. 1. Press APPS #6 "Data/Matrix Editor" #3 "New" Under the "Type" menu, choose option 2 Now you need to choose a name for your matrix, such as jp or jp1 (that is, your initials)

Type the name into Variable: Type 3 in row dimension Type 3 in col dimension ENTER twice Now enter the values in the appropriate rows and columns: 2 ENTER 6 ENTER 2 ENTER 0 ENTER 4 ENTER 2 ENTER 2 ENTER 2

ENTER 4 ENTER 2nd QUIT to go back to homescreen On the entry line, type the name of the matrix (that is jp or whatever you named it), ENTER

You should see 2 6 20 4 22 2 4

.


2. A second way to enter a matrix on the TI-89: Some guidelines: a) Matrices are enclosed in square brackets b) Commas ( , ) are used to separate individual elements of a row c) Semicolons ( ; ) are used to separate rows.

Type the following on the entry line: 2, 6,2;0,4, 2;2,2, 4 ENTER . You should

see2 6 20 4 22 2 4

If you want to refer to this matrix by a name, to save it, for instance, after you have 2, 6,2;0,4, 2;2,2, 4 on the entry line, press the STO key and type the name of the matrix. To find the determinant of your 3 3 matrix: 2nd MATH 4: "Matrix" 2: "det" 2nd Ans 16 To find the cross product:

2nd MATH 4: "Matrix" L: "Vector ops", right arrow and choose 2: crossP( Press ENTER crossP( will appear on entry line. Then type 1,1,3 , 2,2, 1/ 2 ) ENTER

The calculator returns the answer 13/ 2,11/ 2, 4 To find the dot product:

2nd MATH 4: "Matrix" L: "Vector ops", right arrow and choose 3: dotP( Press ENTER dotP( will appear on entry line. Then type 1,1,3 , 2,2, 1/ 2 ) ENTER

The calculator returns the answer 32

.

To delete variable names (names of matrices): Press 2nd VAR-LINK highlight what you want to delete, press F1 ENTER ENTER

CLASSROOM SUPPLEMENT – SECTION 9.6 10

CYLINDERS (also called CYLINDRICAL SURFACES)

Sketch the cylinders by hand and then verify with Maple. Describe the surfaces in words. Example 1: Sketch the cylinder 24z y by hand (on an , ,x y z axis). Describe the surface in words. NOTES ON USING TECHNOLOGY TO GRAPH CYLINDERS To use Maple to graph 24z y , type the equation z 4 y2 , hit ENTER, right click on the blue output equation (control click on MAC), choose plots then 3-D Implicit Plot then ?,y,z since x is not visible in the equation. Maple will give you the graph without axes. To include axes, right click on the graph (control click on MAC), then axes, then choose the type of axes you’d like to see. Normal is the standard you usually see. You may try boxed or framed for a better view. Left click (click on MAC) and hold on the graph to rotate the surface. Simply drag the cursor around. You can also change other features of the graph when you right click (control click on MAC) on the graph. You should play with the available features to find out what style of graphing you like best. To use the TI-89 calculator: MODE At Graph choose option #5 which is 3-D ENTER In Y= you will see z1 = type 4 ^ 2y ENTER In WINDOW, use the following settings: eye =20 min 5x min 5y eye =80 max 5x max 5y eye =0 10xgrid 10ygrid


WINDOW F1 9 allows you to turn axes on/off Choose 2: "axes" to see the , ,x y z axes. Labels: "on" Style: "Hidden Surface" is good (or

"Wire Frame")

Example 2: Sketch the cylinders 1 1 1, , and z z yy x x

by hand.


RECOGNIZING QUADRIC SURFACES

In Exercises 1 - 8, put each equation into standard form for quadric surfaces (for standard forms see page 679 table 2). Then identify the quadric surface as one of the following: cone, ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, hyperbolic paraboloid, or elliptic paraboloid.

1. 2 2 2

19 16 9x y z

5. 2 24 4 0x y z

2. 2 2 215 4 15 4x y z 6. 2 212 3 4z y x 3. 2 2 24 4 4x y z 7. 2 24 4 0x y z 4. 2 2 24 9y x z 8. 2 2 2 9x y z


PARAMETRIZING A PLANE CURVE

Example 1: Find a parameterization for the circle with center (0, 0) and radius 2. Sketch and show the orientation of the circle. Solution:

Example 2: What changes occur (if any) if ( ) 2sin( ) 2cos( )t t t r i j ? Example 3: What changes occur (if any) if ( ) 2cos(3 ) 2sin(3 )t t t r i j ?


SPACE CURVES

Visualizing a space curve. Example 1. Sketch the space curve represented by the vector-valued function

( ) 4cos( ) 4sin( )t t t t r i j k . (i) One way to accomplish this, is to hand-draw the curve on the surface 2 2 16x y .

(ii) Another way is using technology. Directions for using Maple: An easy way to view space curves in Maple is to use its built-in space curve tutor. To access this, click on the Tools tab, then Tutors, then Vector Calculus, then Space Curves as shown below.


The following new window should pop up.

The arrows you see in the graph are tangent, normal, and binormal vectors to the curve at given points. This can be changed as you will see. The space curve is defined in the blue highlighted area above. Below that you can control the range of the parameter t . Notice that within the tutors, multiplication signs (asterisks) must be put in even when it’s not necessary in the original Maple document. Notice that there are five points on the curve for which the vectors are pointing out. This is controlled by the number of frames in the frames box listed under options. If you don’t want to see all three vectors you can choose just one of them in the Display Options box. Try animating the picture by hitting the Animate button. When you do this you can see the trajectory of a particle on the curve through the range of t -values along with the forces the particle would be experiencing according to the force vectors shown. Anytime you alter a parameter in the tutor you must hit Display to see the changes in the graph.


FINDING A POSITION FUNCTION BY INTEGRATION

Class Exercise: An object starts from rest at the point ( 1, – 1, 0 ) and moves with an acceleration of ( ) 2t t a i k where ( )ta is measured in feet per second per second. 1. Find the position function ( )tr . 2. Find the location of the object after 1 second. 3. Use Maple to obtain a graph of ( )tr for 2 2t . Make a careful sketch of the space

curve on this work paper:


VELOCITY AND ACCELERATION VECTORS FOR A SPACE CURVE

Class Exercise: Given that 2

( ) 32tt t t r i j k describes the path of an object moving in space.

1. Sketch, using Maple, the space curve of ( )tr for 2 2t . Draw it below.

2. Find the velocity and acceleration vectors for t = 1. 3. Sketch both vectors on the curve in part 1), at the point where t = 1. 4. Find the speed at t = 1. 5. Find the angle between the vectors (1) and (1)v a . Is the object speeding up or slowing

down at t = 1? How do you know? 6. Find the angle between the vectors ( 1) and ( 1) v a . Is the object speeding up or slowing

down at t = – 1? How do you know?


LEVEL CURVES Graphs and Level Curves The intersection of the horizontal plane z c with the surface ( , )z f x y is called the contour curve of height c on the surface. The projection of this contour curve onto the xy - plane is the level curve ( , )f x y c of the function f. Level curves give us a 2-dimensional way of representing a 3-dimensional surface ( , )z f x y . If we draw the level curves and visualize them being lifted up to the surface at the indicated height, then we can mentally piece together a picture of the graph. The surface is steep where the level curves are close together. The surface is somewhat flatter where the level curves are further apart. Example 1. a) Sketch the paraboloid 2 225z x y on a set of xyz - axes, and

on the paraboloid, show the contour curves for z 0, 9, 16, 21, and 24. b) Sketch the level curves for 2 225z x y on a set of xy - axes.

Practice Example: a) Sketch the level curves for 2( , )f x y x y for c = – 1, 0, 1, 2, 3 (by hand).


b) Use Maple to graph the surface and verify the level curves. Use the following instructions:

There’s a tutor in Maple called Cross Sections (i.e. contour curves). To find it do the following.

This tutor looks like the following.

Enter your function f (x, y) in the expression box. The Plane Equation(s) box on the right puts a range on how low and how high you want the contour curves to go. If you change any parameters in the tutor, hit Display to update the graph. You may also want to animate the picture by hitting Animate. You can rotate the graph like any other graph in Maple.


LIMITS AND CONTINUITY

Recall the definition of limit from Calculus I: lim ( )x a

f x L

if ( ) f x L

whenever 0 x a . We only have to approach a from two directions; the left and right of a. The definition for a function of two variables is:

( , ) ( , )lim ( , )

x y a bf x y L

if ( , ) f x y L whenever 2 20 x a y b . Now

the - neighborhood of ( , )a b is an open disc. The difference between this definition and the definition in Calculus I, is that we must be able to approach ( , )a b from any direction and there are infinitely many such directions. It is extremely difficult to prove a limit exists because we must find a for a given and do a proof based on the definition of limit, just as was done in Calculus I. However, we can show a limit does not exist by finding one "path" or direction toward ( , )a b where the limit either does not exist or has a different value than the limit along another path towards ( , )a b .

Example 1: Show, analytically, that 2 2

2 2( , ) (0,0)lim

x y

x yx y

does not exist.

Solution: We will show that we get two different limits as we approach (0, 0) along two different paths. (i) First approach (0, 0) along the x - axis (that is, the line 0y ):

2 2 2

2 2 2( , ) (0,0) ( ,0) (0,0)lim lim 1

x y x

x y xx y x

(ii) Now approach (0, 0) along the y - axis (that is, the line 0x ):

2 2 2

2 2 2( , ) (0,0) (0, ) (0,0)lim lim 1

x y y

x y yx y y

Because there are two different values for two different approaches to (0, 0), we say that the limit does not exist. Approaching (0, 0) along the x and y axes is a common technique and in this example was enough to show the limit did not exist. Sometimes it is necessary to approach along the line

2 or 2 or even along y x y x y x . For instance, if we approach (0, 0) along y x in this example, we would have:

2 2 2 2

2 2 2 2 2( , ) (0,0) ( , ) (0,0)

0lim lim 02x y x y

x y x xx y x x x

See Table 2 on page 750 in the text for numerical evidence that the limit does not exist. For instance, in the column under x, locate 0 and follow across the row that 0 is in; all the function values are – 1. Look in the y row; locate 0 and look down the column; all the function values are 1.


To get graphical evidence that the limit does not exist, we will use Maple.

Type in z x2 y2

x2 y2 and hit ENTER. Right click on the blue output equation (Ctrl & click on a

MAC) then go to PLOTS then 3-D Implicit Plot then x,y,z. Put in a set of axes as mentioned before. You can rotate the graph as mentioned before and you can also zoom in to see the

discontinuity at (0,0) by clicking on the icon ,then drag the cursor over the graph while holding down the left mouse button like you would with rotating it.

Don’t forget that you can change the range on all three variables if that would help you better see the discontinuity at (0,0) . Right click (control click on MAC) on the graph, go to axes, then properties then uncheck the ‘use data extents’ box and change the min and max to what you’d like. Click on the other variables at the top of the axis properties window to change their data ranges.


Exercise: Show, analytically, that 2 2( , ) (0,0)lim

x y

yx y

does not exist. Then graph this surface with

Maple to support your claim. Example 2: Some limits do exist. Use direct substitution as in Calculus of a Single Variable.

Find 2 2

2 2( , ) (1,2)lim

x y

x yx y

.

Solution: By direct substitution, the answer is 35

.

The definition of continuity in Calculus of a Single Variable, had three conditions, the third of which included the first two, that is , ( )f x is continuous at x a iff lim ( ) ( )

x af x f a

. A

function of two variables is said to be continuous at ( , )a b if ( , ) ( , )

lim ( , ) ( , )x y a b

f x y f a b

. (See

text page 753.) Classifying Continuous Functions: Polynomials and exponentials are continuous for all reals; rational functions, and trigonometric functions are continuous at all points in their domains.

Example 3: Discuss the continuity of 2 22( , ) x yf x y

x y

.

Solution: This is a rational function and therefore, it is continuous at every point in its domain, that is, ( , )f x y is not continuous where 2 2 0x y or 2 2y x , which implies y x . Writing the answer in set notation, ( , )f x y is continuous on , x y y x . Graph this surface with Maple. Can you see the lines y x ? Try rotating the graph a bit.

Exercise: Discuss the continuity of 2 2sin( )( , ) xyf x yx y

.

Classroom Supplement – Section 11.3 23

PARTIAL DERIVATIVES We will use Maple to do Example 2 on page 760 in the textbook. In this example we will visualize the geometric meaning of ( , ) and ( , )x yf x y f x y . Example 2: If 2 2( , ) 4 2f x y x y , find (1,1) and (1,1)x yf f and interpret these numbers as slopes. Maple has a nice tutor that will not only help us visualize the above partial derivatives, but it will also give us the values of the derivatives at (1,1). However, Maple’s tutor is for a generalization of partial derivatives, called directional derivatives. Partial derivatives are special cases of directional derivatives. Carefully follow these instructions to get the partial derivatives: To launch the tutor, click Tools then Tutors then Calculus-Multi-Variable then Directional Derivative. The tutor has a box for entering your function and boxes for entering your x any y coordinates. But then the tutor has two boxes entitled Direction. To get fx the direction must be [1,0]. To get fy the direction must be [0,1]. Hit display once you’ve entered your information. Do not do an animation at this point. Let’s find the value of fx (1,1) and the geometric interpretation of the value of this partial derivative. Here’s what the tutor should look like once you enter the information and hit Display. According to Maple, fx (1,1) 2 . What does this mean? Technically it means that as ‘x’

increases by 1 (while y is fixed), z decreases by 2, since zx

2 21

. How can we see this in

the graph? The pictured plane is the tangent plane to our surface at (1,1,1). (According to the function, z is 1 when x and y are 1). The blue vector is parallel to the xy-plane and is pointing in the direction of the positive x-axis. Notice that the green vector is pointing in the same direction as the blue vector but it lies in the tangent plane. The slope of this vector in the xz-plane is -2, i.e. fx (1,1) . Rotating the graph may help you see this better than it is currently pictured. See also page 760 Figures 4 and 5 in our book for an additional explanation.


THE GRADIENT

CLASS PRACTICE EXERCISE: Let 2 2( , ) 9f x y x y , a) Sketch the graph of f in the first octant and plot the point (1, 2, 4). (This is a 3-dimensional

sketch.) b) Find the gradient ( , )f x y . c) Find the gradient (1, 2)f . Locate the gradient on your sketch in part a). Explain what

information the gradient gives you about the surface at (1, 2, 4). d) Use the gradient vector to find the tangent line to the level curve ( , ) 4f x y at the point

(1, 2). What property of the gradient are you using to do this? Sketch the level curve, the tangent line, and the gradient vector. (This is a 2-dimensional sketch.)

e) Find (1, 2)D fu where 1 3, 2 2

u . Interpret this quantity geometrically (explain what it

means in words). f) Find an equation of the tangent plane to the surface 2 29z x y at (1, 2, 4). Write the equation in both the general form and solved for z that is, as z f (x, y) . g) Using Maple, graph the surface and the tangent plane. First graph the surface using 3D

Implicit Plots, x,y,z. Then to add the tangent plane to that graph, you can enter the equation of the plane in a new maple input. Hit enter, then highlight the blue output equation. Left click and hold while dragging the equation onto the graph of the surface. Release and Maple will automatically graph the plane with the surface.


FINDING EXTREMA OF MULTIVARIABLE FUNCTIONS

Example: Find the extrema of 4 4( , ) 4f x y xy x y . Solution: First find and x yf f and then solve for the values of x and y that make both and x yf f

equal to zero, at the same time. 34 4xf y x =0 3 y x . Now substitute 3y x into yf set equal to 0.

34 4yf x y =0 9 4 4 0x x or factoring, 84 (1 ) 0x x or x = 0, 1, – 1. Therefore, the critical points are (0, 0) (1, 1) (– 1, – 1). Test each critical point using the Second Derivatives Test (for Functions of Two Variables): First find , , xx yy xyf f f a) At the critical point (0, 0):

D

b) At the critical point (1, 1):

D

c) At the critical point (– 1, – 1):

D

Use Maple to produce a graph of 4 4( , ) 4f x y xy x y and 8 of its level curves, using the Cross Sections Tutor (see pg 19 of this supplement for instructions to do so). For the contour curves, use z -values of 0,1,2,3, 4 , i.e. change the Plane equations to z 3...4 and the number of planes to 8. Do the contour curves confirm your analytical test? Try animating this one. (See text explanation for figure 5 on page 804; this is a different example but the ideas are the same.)


Iterated Integrals, using technology

(First do the following example by hand, using the order dydx , and then verify using technology.)

Example: Evaluate 3 4 2 3

1 1x y dydx

. It’s rather easy to enter this iterated integral into Maple.

In the Expression palette, click on . To make this a double integral, tab to the integrand

position denoted by f and click on again. This creates the double integral template,

. Enter your bounds for both integrals, your integrand and be sure to enter the

appropriate order of integration. You should get x2y3 dydx 5951

4

1

3

.

Do the same problem with the order dxdy . You should get the same answer. ______________________________________________________________________________ To use the TI-89/92: Press the "second" and integrate keys: 2nd

When the key is pressed, a left parenthesis automatically follows the integral sign; then type the integral as follows: 2nd x^2y^3, y, 1, 4), x, 1 , 3) ENTER Answer: 595


USING Maple TO EVALUATE DOUBLE INTEGRALS

Example:

1 1 2 2

0( )

yx xy y dxdy . You enter this double integral into Maple, the same way

you do with iterated integrals, mentioned on the previous page of this supplement. Please don’t forget that the xy term must be entered as x y . Otherwise Maple will treat xy as a new

variable name. Once entered, Maple returns the answer 33

140.

______________________________________________________________________________ To use the TI-89/92: Press the "second" and integrate keys: 2nd

When the key is pressed, a left parenthesis automatically follows the integral sign; then type the integral as follows: 2nd x^2 + x*y y^2, x, (y) , 1), y, 0, 1) ENTER Answer above.


REVERSING the ORDER of INTEGRATION

To Determine the Limits of Integration for Double Integrals 1. Make a quick sketch of the region. 2. Draw an approximating rectangle on the sketch, either horizontal or vertical. 3. The width (shortest side) should correspond to the outer variable of integration. 4. The height (for a vertical rectangle) or length (for a horizontal rectangle) will determine the inner

limits of integration. The height is determined by top value minus bottom value, or the length is determined by right value minus left value.

5. Choose the type of rectangle that results in the simplest integral.

Class Practice Exercise: Given the double integral 1 1

20

11y

dxdyy ,

a) Evaluate this integral, by hand. b) Sketch the region D of integration, that is the region in the xy - plane given by the limits of

integration. c) Use your sketch to reverse the order of integration, that is, set up the integral

21

1dydx

y .

d) Start to evaluate this integral by hand. Is this integral easier or harder to do by hand than the one in step a)?

e) Use Maple to integrate the integral you set up in part c).


USING POLAR COORDINATES TO FIND THE VOLUME OF A SOLID

Example 1: Use polar coordinates to find the volume of the solid bounded by the paraboloid

2 210 3 3z x y and the plane 4z . Use Maple to help visualize the solid. Remember, graph z 10 3x2 3y2 then add the plane z 4 to the graph by entering z 4 in a new maple input, followed by highlighting the blue output equation and dragging it onto the graph of z 10 3x2 3y2 . Sketch the solid here: What is the curve of intersection of the two surfaces? The integrand is (top surface – bottom surface):

2 2 2 210 3 3 4 6 3 3x y x y = 2 26 3( )x y = 26 3r The double integral in polar coordinates is:

2 2 2

0 06 3 6r rdrd

(Verify the integration by hand).


Example 2: Use polar coordinates to find the volume of the solid bounded by the paraboloids

2 2 2 23 3 and 4z x y z x y . Use Maple to help visualize the solid or sketch the solid yourself. Answer: 2


PLOTTING VECTOR FIELDS USING MAPLE Maple has a built in vector field tutor. Go to TOOLS then TUTORS then VECTOR CALCULUS then VECTOR FIELDS. You should get a screen such as this one below.

The commands in the Tutor are similar to all the other ones we’ve used, except for perhaps Initial Point. By defining an initial point, you can see how a particle will be affected by (move under) the force field. The particle’s trajectory is the space curve in the graph. Of course this is a 3-D vector field. You can reduce the vector field by a component to get a 2-D vector field. The Display button will trace out a particle’s movement along its trajectory from the defined initial point.

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GRAPHING POLAR EQUATIONS