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Application of integrals

  1. Sep 2, 2005 #1
    Hi everyone. I would really appreciate it if someone could check my work. I was really unsure on how to do some of these. I'm sorry it is so long.

    1.) Let R denote the region between the curves y=x^-1 and y=x^-2 over the interval 1<= x <= 10.

    a. Set up an integral for the area of R.
    My answer: 1.403

    b. Find x-bar, the x coordinate of the centroid of R.
    My answer: 4.775

    c. Set up and evaluate an integral for the volume of revolution of the solid generated when R is revolved about

    i. the x-axis
    My answer: 1.781

    ii. the y-axis
    My answer: infinity

    2.) The length of a cable is 50 and the weight is 10. A portion of length 40 was hanging over the edge of a tall building and was pulled to the top. How much work was done?
    My answer: 3920

    3.) Let C denote the curve y= x(4-x), where 0<= x <= 4. Set up the integral for the following. In this case, do not evaluate the integrals.

    a. the length of C
    My answer: integral from 0 to 4 of sqrt[ 1+ (4-2x)^2] dx

    b. the area of the surface generated when C is revolved about

    i. the x-axis
    My answer: 2pi *integral from 0 to 4 of x(4-x)*sqrt[1+ (4-2x)^2] dx

    ii. the y-axis
    My answer: 2pi* integral from 0 to 4 of [sqrt(4-y) -2] *sqrt[1+ (-2*sqrt(4-y))^-2] dy

    4.) A tank has the shape of a trapezoidal prism. The top is horizontal and the two ends are vertical. The length is 4. The height is 2. The top is a 3-by-4 rectangle. Viewed from an end, the tank looks like the trapezoid shown in the figure below. Assume the tank contains a liquid to a depth of 1. Take the density of the liquid to be p.

    a. Set up, but do not evaluate, an integral for the work required to pump the liquid to the top of the tank.
    My answer: W= p*integral from 0 to 2 of 4 dy

    b. Set up, but for not evaluate, an integral for the fluid force against one end of the tank.
    My answer: F= integral from 0 to 4 of p(24) dx

    http://img361.imageshack.us/img361/559/calctest3nq.png

    Thank you
     
    Last edited: Sep 2, 2005
  2. jcsd
  3. Sep 2, 2005 #2

    Hurkyl

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    It's hard to check your work when you don't give it!

    By the way, you really ought to give exact answers, rather than decimal approximations.
     
  4. Sep 2, 2005 #3
  5. Sep 3, 2005 #4

    HallsofIvy

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    But the problem didn't ask for the area! It asked you to set up the integral. Your answer should be the integral, not a number! Some teachers get very sticky about that!
     
  6. Sep 3, 2005 #5
    I'm sorry it said "set up and evaluate."
     
  7. Sep 3, 2005 #6
    For #1 all is correct except for the infinite volume an part c(ii). I used the other volume method [ integral from x=a to x=b of 2*pi*x*f(x)*dx]. In certain problems (usually those where you are revolving around the Y axis) this method is a lot easier.

    If you prefer the method that you tried then I see why you got infinity. You have the region bounded by the two functions x=1/y and x=1/sqrt(y). It should also be bounded by a third function, x=10. Take this into account and you should get a finite answer.

    For #2 it looks as if you multiplied the W=weight by the acceleration due gravity (the 9.8) and integrated. What you should do is find the weight density of the rope (p=W/L where W=total weight and L=Total length). Now cut the rope into small intervals. You know that the weight of any interval is Δw=p*Δl or dw=p*dl. You now that work is the integral of force*distance, you know know the force (dw) and you know the distance the section of the rope has to be lifted (the distance between the section of the rope and the top of the building.

    I believe part three is correct but I am mostly used to doing surface area for parametric equations. Mabye someone else reading this will double check this part..............

    For #4 you should do something similar to what I told you to do in #2. In fact it is so similar to #2 that I believe that it will come easy to you after you redo #2.

    I realize that sometimes when I write I am not as clear as possible. If I was not clear about something than just post a reply asking for more explanations.
     
    Last edited: Sep 3, 2005
  8. Sep 3, 2005 #7
    Is the answer to number 4:
    W= Int y 10/50dy= 10/50 y^2/2 limits 0,40 ? and I'm sorry but I still don't really understand how to do number 4
     
  9. Sep 4, 2005 #8
    There are many different ways this integral can be written so you should show your work so I can see how you got the integral. However, I dont think the answer you got is correct for #4.

    Below is some tips on solving the problem. I tried to give you as much help as possible without just giving you the answer so that you can have the fun of solving it yourself. Again, if I was not completely clear in my explaination then just tell me and I will give more info.

    My recommendation would be to use this as guide to solving the problem and then practice more problems like it.

    It was easier for me to write out the explanation and then scan instead of writing the explanation out because the explanation includes drawings and stuff. The problems is that I am not good at using this scanner so the page is very big and you will have to scroll around to see the whole explanation. Sorry for the incovinience.

    http://gozips.uakron.edu/~lm24/
     
    Last edited: Sep 4, 2005
  10. Sep 5, 2005 #9
    So what would the integrals for number 4 be?
     
  11. Sep 7, 2005 #10
    Sorry for taking so long to reply to your last post. From now on I will be quicker when I am supposed to be helping somebody with something

    Usually on this site we dont give answers to problems unless we know you already worked very hard on the problems. I can tell that you have so here is the integral to #4 part I

    [tex]\int_{0}^{1}4p(h+1)(2-h) dh[/tex]

    First note the form of the integral. Work is usually written as [tex]\int fdx[/tex]. In this equation the distance is in the form of a differential dx. In my equation its more like [tex]\int xdf[/tex]. The reason is this: in this problems the distance that a layer of water at height h has to travel is simply x=(2-h). However, the layers of water are infinetely small, because the h axis was divided into peices and become infinitely small just like in normal integration the x axis is divided into infinetely small parts. Since the height is infinitely small then the volume of a layer is infinetely small. The weight of a layer (which is also the force required to lift a layer) depends on volume by the equation pV=F where F is force or weight and p is weight density, defined is Weight/volume, and V is volume. If the volume is infinetely small then so is the force and therefore the force is dF. The layers of water are rectangular with dimensions w, where w is the width of the layer (w is the lenght between the two sides of the trapezoids, as can be seen in my diagram in the link I posted earlier. since the sides of the trapezoids are lines that increase with height then the widht is a linear function of height. in this problem the width is 1 when the h is 0 and w is 3 when h is 2. the only linear function satifying these two conditions is w=h+1), dh, which is the height of the layer (notice that the height is a differential. this is because we divided h into sections Δh and when we take the limit it becomes the differential dh) , and 4, the length of the tank. Volume is lenght times width times height or 4(h+1)dh. If thats the volume then the weight is df=4p(h+1)dh. We have force and we have distance. multiplying them together and rearanging gives 4p(h+1)(2-h)dh. This is the work requred to lift an individual layer. To pump out all the water we have to sum the work required to lift every layer. The sum goes from h=0 to h=1 since the water goes from h=0 to h=1. The integral is then

    [tex]\int_{0}^{1}4p(h+1)(2-h) dh[/tex]

    refer to the link I posted earlier for a little more detail on some of the steps

    remember, there are different ways of doing this so if you come up with a different answer it may still be right. If you want to check, you know your answer is right if you evaluate your integral and get the same thing you would get if you evaluated my integral.

    For #2 I used j for depth in this post instead of d. Notice that in the link I posted earlier, that I used a different coordinate in parts I and II; I used d (which is now j) for depth, instead of h, and I positioned this axis differently than the h axis. This is just to make the problem a little simpler. If you prefer you could use h on both problems or d (again, is now refered to as j) on both. Of course this would make your integral a little different but like I said, there are many different right answers

    My final answer was

    [tex]\int_{0}^{1}p j (2-j) dj[/tex]

    I started with the formula F=paj where F=Force p is defined as weight/volume, a=area, and j=depth. the j axis is divided into layers Δj and after taking the limit becomes dj. The dimensions of each layer are w, which is width and dj, which is height. We dont need the length=4 in this problems because we are concerned with area, not volume. The area is then w*dj. Using logic similar to what I used to find that w=h+1 in the first problem, I find here that w=2-j. then the area is (2-j)dj. We now can use the formula Force=paj=p(2-j)jdj. This is the force on an infinitely small section of the trapezoid. For the total force we have to sum over j. The sum goes from j=0 to j=1 since the water goes from j=0 to j=1. the final integral is then

    [tex]\int_{0}^{1}p j (2-j) dj[/tex]

    I hope I was clear enough in my explanation for you to understand it. If not just ask for more info (and if you do ask for more info I promise to reply quicker this time) :biggrin:

    My recommendation would be to practice more problems like this until these problems seem extremely easy to you. Then you can come to physicsforums and help other people out :smile:
     
    Last edited: Sep 8, 2005
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