- By
**naida** - Follow User

- 232 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'CHAPTER SIX: APPLICATIONS OF THE INTEGRAL' - naida

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### CHAPTER SIX:APPLICATIONS OF THE INTEGRAL

Hare Krsna Hare Krsna Krsna Krsna Hare Hare

Hare Rama Hare Rama Rama Rama Hare Hare

Jaya Sri Sri Radha Vijnanasevara (Lord Krsna, the King of Math and Science)

KRSNA CALCULUS™ PRESENTS:

Released by Krsna Dhenu

February 03, 2002

Edited on October 7, 2003

HARI BOL! WELCOME BACK!

- Jaya Srila Prabhupada! Jaya Sri Sri Gaura Nitai! Jaya Sri Sri Radha Vijnanesvara!
- I hope you are spiritually and materially healthy. This chapter is where we start to depart the AB calculus students and College Calculus I.
- If you take AB Calculus or Calculus I, then I urge you to please check out my very very final slide show. Thank you and much love.
- If you are a BC student, however, this is your middle point of the course. For Calculus II students, welcome to Krsna Calculus ®. I will enjoy having you as my students.
- Calculus II entrants: please check out chapter 4 and 5 extensively.
- Everyone: This chapter deals highly with other fancy things you could do with the integral. First, finding general area, and techniques will be discussed. Then the big part of the chapter on volume of a solid of revolution. This may seem hard, but it is really not.
- Other applications include arc length AND surface area.
- When we start talking about cylindrical shells, AB and Calculus I students are completed with this course and are urge to go and take a review slide show for them.

AREA

- In chapter 4, we introduced integration in terms of area between the curve and the x-axis.
- This chapter, we will discuss area between two curves and methods.
- Also, we will talk about integral and derivative properties in terms of their graphs.

AREA BETWEEN TWO CURVES

- The area between two curves is really simple if you really look at it. Let’s consider f(x)=6-x2. Also consider g(x)=x.
- Find the area between these two curves.
- Here is a graphical view of the scenario. (Magenta region) ALWAYS DRAW A PICTURE TO SEE HOW THE REGION WILL LOOK LIKE!!!!!!!

STEP 1: FIND THE POINTS OF INTERSECTION

- In order to see where the region “begins” and “ends,” we must find the x limits to see where we can actually compute area.
- Solve for the x limits
- The two limits of integration are x=2 and x=-3

GEOMETRICALLY

- This is a shaded region style problem. Let the maroon be the region between the parabola and the x axis. with the limits of -3 and 2. Let the light blue be the region between the line and the x axis with the same limits. Let the purple represent overlapping regions.
- We must subtract the two areas of the regions (that is Af(x)-Ag(x)) to get rid of the overlapping areas (i.e. purple region).

If f(x)>g(x)… top vs. bottom function???

TOP

- Very simple..
- Since the parabola, f(x), is on top of the line, g(x), therefore, f(x)>g(x).

BOTTOM

AREA BETWEEN TWO CURVES

- We just tackled one of the three types of area between curves problems. They are
- 1) Area between two y(x) functions. (We just did)
- 2) Area between a y(x) and x(y)
- 3) Area between two x(y) functions.

AREA BY SUBDIVIDING

- Sometimes, it is necessary to split the region up into two parts before integrating. This situation often occurs when you have y as a function of x and x as a function of y.
- Consider this example: Find the region bounded by y=½x and x=-y2+8.
- First thing you do is to draw a picture to see how the region will look like. Then solve for their intersecting points.

DRAW A PICTURE!!!

- It helps to draw a picture to get an understanding on what is going on. The red line is y=½x and the blue parabola is x=-y2+8.

INTEGRATING

- It is very important to let all the functions be in the form y(x). Therefore, solve for y.
- Using the formula given previously for finding the area between curves, apply that formula using -8 and 4 as your limits.

WHAT AREA DID WE JUST DO???

- Notice how much area we did accumulate using the integral.
- Also consider how much area we have left to take into account.
- The purple is the area we jus did. (76/3) The green is the area we have left to do.

THE GREEN AREA

- Since we know that the area of the green is the area of that sideways parabola, and since we also know that the area above the x-axis from [4,8] equal to the area below the x-axis. So in effect, we can double the area of that to take into account the top and the bottom of the x-axis.

CALCULATION OF AREA

- Notice the 2 outside the integral. We have to double the area.
- Fundamental theorem.
- Add the two areas.
- THE TOTAL AREA OF THE REGION IS 36!!

INTEGRATING A FUNCTION WITH RESPECT TO Y.

- At times, we may have to curves that there is no “top” or “bottom” y, rather there are left and right curves. By definition, these curves would not be functions of x, since they would fail to honor the vertical line test (the test that determines if a relation is actually a function). But these functions, on the hand, seem simpler if they were in terms of y.

INTEGRATING WITH RESPECT TO Y.

- To calculate an area, using the right being x=f(y) and the left one being x=g(y) function, between limits of y (c and d)…then…

OR

EXAMPLE

- Let’s do the previous problem using this method of integrating with respect to y.
- The graphs are y=½x and x=-y2+8.
- First, always always always!! Look at a picture of this! This will help you!
- The red is the line. The blue is the parabola. The purple is the region.

RIGHT

LEFT

SOLVE FOR INTERSECTING POINTS

SAME SLIDE AS BEFORE, HOWEVER, CONSIDER THAT THE Y VALUES ARE NOW MORE RELEVANT, SINCE WE ARE INTEGRATING WITH RESPECT TO Y!!!!

THIS MEANS THE INTEGRAL WILL HAVE THE LIMITS BETWEEN y=-4 AND y=2!!!

SETTING UP THE INTEGRAL

- First solve everything for x before putting the f(y) and g(y) inside. You will get x=2y AND x=8-y2.

Simpler, eh???

AREA BETWEEN TWO CURVES

- In reality, there is no set formula to finding area between curves. It’s important to know what to do and how you go about doing it. For example, you noticed that integrating a function in terms of y was simpler than taking the region, cutting it, and finding areas that way.
- But sometimes, you will see that integrating with respect to y can be a hassle. Sometimes, integrating with x would be better.
- But in either case, the general formula stays the same: top/right – bottom/left. (for x and y respectively)

EXAMINING THE INTEGRAL

- A calculus course is not only about, “Do you know how to calculate the integral.” However, this course also asks you, “Do you know how to find the integral qualitatively?”
- Consider this graph of g(x). The function f(x) is its derivative. Therefore f’(x) = g(x).

PROPERTIES OF THE GRAPH

- Lets talk about limits. Notice at x=1, the limit from the left side is a little more than 5, but from the left side, its 0. Therefore, the limit at x=1 does not exist.
- g(x) is not differentiable at some point between -6 and -7 and at -2, since they cause vertical tangents. Also x=-2 is not differentiable, since it is a corner. x=-1 is also a corner, thus it is not differentiable either. x=1 is not differentiable, since it is a discontinuity. Lastly, x=5 cannot be differentiated since it is the end point of the graph.
- g’(x) is positive at approximately (-6.5,-4.3), (-1,1), and (2.75,5).
- g’(x) is negative at (-4.3,-2) and (1,2.75).
- g’(x) is zero with maxima at x=-4.3 and minima at x=2.75.
- f(x) is the area between the curve and the x-axis. Thus, the curve of f(x) is the antiderivative. Since g(x) is the slope for f(x), this means that f(x) is increasing slowly, and maintains an equilibrium from x=-2 to x=-1. From x=-1, the the function increases like a parabola, but due to the discontinuity of g(x), a cusp was formed so that the slope is negative throughout.
- REMEMBER!! THE SIGN OF f’(x) OR g(x) IS THE SIGN OF THE SLOPE!!!!

INTEGRAL MEANS AREA ACCUMULATION

- Look at the left graph f(x)=1/x and the right graph g(x)=ln x. Obviously, g’(x)=f(x). Notice how the 1/x decreases very quickly, but nevertheless, adds more area. Although not at a fast RATE, but it still ACCUMULATES area!

VOLUME OF A SOLID OF REVOLUTION

- FOR AP CALCULUS AB STUDENTS THIS IS YOUR FINAL TOPIC
- FOR CALCULUS I: YOU ARE COMPLETED FOR THE COURSE. YOU ARE WELCOMED TO STAY, BUT YOU CAN LEAVE IF YOU WISH. TAKE CARE! BEST WISHES! HARE KRSNA!
- FOR CALCULUS II STUDENTS: HARI BOL! WELCOME! THIS IS YOUR FIRST TOPIC FOR THIS CLASS!!!!

VOLUME

- Volume and 2-D calculus?? Possible?
- Yeah, I guess.
- Imagine if you were to take a region between f(x), the limits of integration, and the x-axis, and you were to swing that region about the x-axis.
- Take your hand and try that. Put your hand on the region, and turn your hand such that you rotate about the x-axis.

CLARIFICATION

- What we did was we took the region under the line from x=0 to x=7 and rotated that region 360 about the x-axis.
- You will notice that a solid is formed. That solid looks like a cone in this example.
- My computer does not really have great 3-d effects so please forgive my drawings.
- How do we find the volume of this cone using calculus?

TORICELLI’S LAW

- An Italian mathematician suggested that you can find the volume by take adding the areas of the cross sections.
- He also proved that the volume for two geometric objects, despite “slant”, would still be equal. Take a look at these two cylinders. Although it looks very obscure on the right, it will still have the same volume as the cylinder on the left.

THEREFORE:

- If A(x) is the cross section for such a volume, and you could get a Reimann sum saying that the width of the solid multiplied by the area of the cross section…

IF THE WIDTH

- If the width was infinitesimally small as dx, then the area would represented as an integral.

CUTTING UP THE SOLID

- Imagine if we were to cut up the solid to infinity slices! That would that the thickness of the slice would be virtually 0. But let’s take out one of these slices …

dx

THE SLICE

- The cross section of such volume is a circle. The radius of the circle would merely be f(x) value, since the center of the circle is the x-axis. Look at the 2-D and the 3-D illustration.

The radius of the cirlce

is f(x).

r=f(x)

x axis

AREA OF A CIRCLE?

- Let’s say that the line was the function
- f(x)= ½ x
- Remember that the integral of the area of the cross sections from a to b is the volume? Also remember that f(x) is the radius?
- If the circle is the cross-section for the solid, then what is the area of a circle?
- C’mon! Don’t tell me you are in calculus and don’t know the area of a circle!
- A=pr2 ! No! pies are round!

If the radius and the function are equal, so are their squares

- f(x)=½x
- Apply the squared radius into the area of circle expression
- Volume-as-an-integral expression
- Limits of the original problem were x=0 and x=7.
- Evaluation of the integral. The volume is found

RECALL

- We took a region under f(x)=½x from x=0 to x=7 and rotated it around the x- axis to form a solid. We took this solid and cut it up to infinite slices. We took one of those slices and examined that the cross-section is a circle (most commonly called a disk). We found the area formula for the disk. A=pr2. Since r=f(x), we put it into the formula. The volume formula says that we take the integral from 0 to 7, for the [f(x)]2. That result yields into the volume.

DISK METHOD FORMULA

- If the cross sections are circles, and if region are is being rotated around the x- axis, then the formula for finding volume is:
- If the region between g(y) and x -axis is being rotated about the y-axis, with disk cross section, then the area formula is this.
- Don’t forget to put the p in there.

DISK METHOD

- Remember how we had instances that we were forced to integrate with respect to y?
- If that is the case, solve for y, use the y-values for the limits of integration, and use the same formula.

WASHER METHOD

- Find the volume of the region between y2=x, and y=x3, which intersect at (0,0) and (1,1), revolved about the x–axis.
- First, DRAW A PICTURE

THE SOLID

- Notice that the solid now has a “hole” inside it (very light blue). The edges of this solid (darker light blue) (is now created by two functions.
- Since we are revolving this solid around the x-axis, we must solve everything in terms of x.

SHELLS FORMULA

- If we solve for x, then we will get √x. (square root) and x3.
- If you look at the cross section, you will get something similar but somewhat different from a disk. In order to take into account, the hole in the bottom, we have to include a hole inside the disk. Therefore, the slice will look like this.

g(x) BOTTOM FUNCTION

f(x) TOP FUNCTION

THE AREA OF THE ORANGE

- If you wanted to find the area of the orange portion, then find the area of the entire circle and subtract the area of the inner cirlce. A typical shaded region problem! Capital R means outer radius, and lower case r means “inner radius.”

VOLUME

- Having infinite “doughnuts” that are very thin, you can use the integral to find the volume. This is called WASHER’S FORMULA.

R and r

- If you also consider it, the outer radius is the top function while the inner radius is the bottom function, since the bottom function borders the hole in the center of the “doughnut.”
- Let’s call this “doughnut” a washer from now on!!!!
- I don’t wanna make ya guys hungry yet!
- (15 mins. Later) Now I’m hungry!!!!

FINDING THE VOLUME

- Wahser’s formula
- DON’T FORGET THE p!
- Original limits and top and bottom functions are included.
- The final volume evaluated.

INTEGRATING w/ RESPECT TO Y.

- Let’s take the previous problem and rotate it about the y-axis.
- That means, we will have R as right function and r as the left function
- Therefore: x=y2 and x=y⅓.
- ALWAYS DRAW A BEFORE/AFTER PICTURE!!!

LINE ROTATIONS

- Suppose we take the previous example and rotated it about the following lines
- A) x=-1
- B) y=-1
- Find both volumes.

DRAW A PICTURE

- Draw both before/after diagrams for the x=-1 rotation and y=-1 rotation.
- Please forgive the cut-offs in my drawings.

THE SHIFT!!!

- Its not a movie. LOL
- Look at the reference line (reference line is the line you are rotating the region about.)
- Actually draw out the radii. For the x=-1 rotation, you will need to add 1 to both functions to take into account.
- General Rule: If rotation around a line x=k, then f(x)-k is the big radius and g(x)-k is the small radius.

THE RADII

- Therefore, the radii is 1+y2 and 1+y1/3.
- Since by this time, you are getting the fundamental theorem of calculus down pretty well (or at least you should be), I will stop showing the plugging in of numbers. I will show the integral, substitution, actual integration and answer.

ROTATING about y=-1

- Remember! If you have y=k as line of rotation, then you will f(x)-k and g(x)-k
- Therefore, you will have f(x)=x1/2 +1 and g(x)=x3+1
- Using the washers formula, let’s find the volume.

KRSNACULULUS® STATUS

- END OF AP CALCULUS AB
- END OF CALCULUS I
- MIDDLE POINT OF AP CALCULUS BC.
- START OF CALCULUS II.
- ** AP CALCULUS AB AND CALCULUS I, PLEASE GO TO THE SUMMARY SECTION.**

VOLUME BY SHELLS

- If we wanted revolve a region to form a solid, we could use really thin cylinders, find the area for all of these cylinders and thus, get the volume.
- It is merely the sum of the surface area of each cylinder. The formula of the surface area of a cylinder is 2prh. If we make the thickness as thick as dx, then the shells formula would like this.

ARC LENGTH AND SURFACE AREA

- This topic requires very strong algebra skills. We will introduce arc length and surface area of a solid of revolution, since their formulas are very much similar.
- First, we must discuss “ds.”
- WHAT IS “ds?”

PYTHAGOREAN THEOREM

- Remember the Pythagorean theorem?
- x2+y2=r2
- Let’s call they hypotenuse as “s” instead of “r.”

s

y

x

SIMILARLY…

- Works with differentials!
- ds2=dx2+dy2
- Same concept applies
- You could rewrite it as ds2=[1+(dy/dx)2]dx2

ds

dy

dx

DISTANCE

- Notice if you make right triangles with equal ∆x, the hypotenuse will be really close to the graph line. The hypotenuse would be ∆s.
- The sum of ∆s will give you the distance or the length of the line.
- If we took infinitesimal right triangles such that ∆x would become the width, dx. Therefore, ∆s will become ds.
- The sum of these very minute lines or “dots” (hehe) would become the length of the line.

REIMANN SUM AND INTEGRAL FOR ARC LENGTH

- Definition for ds.
- Reimann sum to find the distance of the line
- THE ARC LEGNTH FORMULA with respect to x.
- THE ARC LENGTH FORMULA with respect to y.

COMPUTATING ARC LENGTH

- Due to the radical sign in the ds expression, integration becomes really difficult.
- In the later chapters, we will discuss integration of such integrals.
- We will only do very simple functions that we can integrate with the knowledge we have now.

FROM THIS FORMULA

- You can get surface area.

SAMPLE PROBLEM

- Let’s say you are given the great honor to make garland for the Deities. Of course, you don’t want them to be so short that the garlands become a necklace. But then again, you don’t want to make them as long as they would touch the floor! In other words, you gotta make ‘em just right!
- Make this a bonafide problem as much as possible. If we were to “break the loop and lie it along the “tilak graph” (y = (2/3)x3/2), you’ll see that the graph starts from x=3 to x = 15. Find the length of this garland.

SAMPLE PROBLEM

- You guessed it! Arc length formula!
- You differentiate the function and square the derivative as shown.
- Put the formula to some action!!
- Oops! Don’t forget u-substitution. If u=1+x, then du/dx = 1, therefore du = dx.
- Using the fundamental theorem of calculus along with the u-substitution, evaluate the integral!

SURFACE AREA PROBLEM

- Remember in the Mahabharata, the five Pandavas needed to be in disguise during their thirteenth year of exile! Arjuna is gonna be Brhanalla, and he needs a pakhwaj drum.
- You are his personal assistant who makes these drum. It’s almost cylindrical but let’s say that is 100% cylindrical.
- Theoretically, the pakhawaj resembles the solid formed by rotating the region over the x-axis. The region is defined as being bound between the line y=3, x-axis, y axis and x=10.
- Find the surface area of this pakhawaj.

GRAPH, BEFORE, AFTER

- Always know what region you are talking about. Draw a graph!! It helps!
- Eh.. I know.. Not the best pic… but you get the idea of how it looks like when rotated!

FINDING THE SURFACE AREA

- Remember! Formula first! Don’t forget! We rotated about the x-axis, so we use the A(x) formula.
- The function is y=3. dy/dx = 0.
- The integral becomes a piece of cake
- Simply apply fundamental theorem of calculus.
- Don’t forget. Area = units sqaured!

END OF CHAPTER SIX

- Considering this the end of Calculus I (AP Calculus AB) students and the beginning of Calculus II (mid-AP Calc BC), everyone should keep the following in mind.
- Don’t forget the most important formulas:

Area between 2 curves

- (all)
- Disk method
- (AB II)
- Washer’s method
- (AB II)
- Cylindrical Shells
- (BC II)
- Arc Length
- (BC II)
- Surface Area
- (BC II)

SUMMARY

- When doing these problems with area, volume, arc length, or surface area, always draw a picture of the all graphs and conditions involved.
- Then draw a before/after scenario.
- Write the equations necessary to solve the problem.
- Check with a calculator if available

CONCLUSION

- We are about to reach a point in calculus where simply memorizing equations and plugging numbers will not be useful all on its own.
- Calculus requires critical thinking. The secret behind math is planning. How do you plan to solve a problem? With method would work the best and efficiently? These questions will come up soon.
- Never forget the important rules of integration. The next few chapters are strictly devoted to integration.

END OF CHAPTER SIX

- Sri Krsna Caitanya Prabhu Nityananda
- Sri Advaita Gadadhara Srivasadi Gaura Bhakta Vrnda
- Hare Krsna Hare Krsna Krsna Krsna Hare Hare
- Hare Rama Hare Rama Rama Rama Hare Hare

CREDITS

- Mr. J. Trapani
- Mr. G. Chomiak
- Dr. A. Moslow
- Mr. D. Ireland
- Calculus and Early Transcendental Functions 5th Ed.
- Single-variable Calculus (SUNY Buffalo)

Download Presentation

Connecting to Server..