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Finding the area of revolution

  1. Dec 2, 2009 #1
    How can i get the volume of this problem


    1. The problem statement, all variables and given/known data

    y = cosx, y = sinx , x = pie/2 , and the y axis finding the volume about x & y axis
     
  2. jcsd
  3. Dec 2, 2009 #2

    Mark44

    Staff: Mentor

    If you want help, you need to show what you have tried to do.

    Also, your problem description suggests that maybe you are trying to find two different volumes of revolution (and not the area of revolution, as is mistakenly stated in your title).
     
  4. Dec 2, 2009 #3
    sorry for the mistake. It should be volume revolving about x and y axis to be found separately.

    I have done graphing all of the equations in the coordinate axis it gives me a square like ( y = sinx , y = cosx upto pi/2 and x= pi/2 and the y axis )shape. Then which of the function i will use to find out my volume ?
     
    Last edited: Dec 2, 2009
  5. Dec 2, 2009 #4

    Mark44

    Staff: Mentor

    You can start by giving us the exact problem description as stated in your book or wherever you got the problem.

    The region that is going to be revolved around each axis (separately) is not a square.

    BTW, this Greek letter, [itex]\pi[/itex], is pi, not pie.
     
  6. Dec 2, 2009 #5
    The exact problem is :

    Find the region enclosed by y = sinx, y = cosx , x= pi/2 and y axis.

    Hence find the the volume of the solid of revolution obtained by rotating the region

    1. About Y-axis

    2. About X-axis
     
  7. Dec 2, 2009 #6

    Mark44

    Staff: Mentor

    OK, the description still seems vague to me, so I believe the region is bounded by the graphs of y = sinx, y = cosx, and the x-axis, between x = 0 and x = pi/2.

    This region has a sort of triangular shape but with two curved sides, with vertices at (0, 0), (pi/4, sqrt(2)/2, and (pi/2, 0).

    If you revolve this region around the y-axis, you get a sort of donut-shaped solid. There are two ways to get the typical volume element: concentric shells and disks.

    By concentric shells, the typical volume element is [itex]\Delta[/itex]V = pi*x^2*f(x)*[itex]\Delta[/itex]x. f(x) represents the height of the shell, which is sin(x) on the first part of the interval and cos(x) on the second part of the interval. This means you will need to use two separate integrals, one for each subinterval, to represent the volume.

    By disks, the typical volume element is [itex]\Delta[/itex]V = pi*(R^2 - r^2)[itex]\Delta[/itex]y. R represents the radius (the x-value) from the y-axis to curve that is farther away, and r represents the radius (the x-value) from the y-axis to curve that is nearer. Since the thickness of the disk is [itex]\Delta[/itex]y, the two x values for the radii need to be written in terms of y, so you'll need to use inverse functions. IOW since y = cos(x) for the more distant curve, x = cos-1(y) gives the radius R. It's going to be much more complicated to use disks, so I would recommend using shells for this part of the problem.

    To find the volume when the region is rotated around the x-axis, use similar reasoning to determine the typical volume element, which can be gotten using the same two techniques I already used.
     
  8. Dec 2, 2009 #7
    I m sure that the region is enclosed by y = sinx, y = cosx , x= pi/2 and y axis.
    It is Y- Axis not X- Axis. Thats why i m confused about this math.

    Thanks
     
  9. Dec 3, 2009 #8

    Mark44

    Staff: Mentor

    The y-axis is x = 0.
    IMO this problem is very poorly worded, because it does not describe the region to be rotated in an unambiguous fashion. Your choices are to:
    1. Get clarification from your instructor.
    2. Make an assumption (and state it explicitly) about what you believe the problem is asking, and continue on from there.
     
  10. Dec 3, 2009 #9
    What clarification do i need from my instructor ? This is the given problem and i m asked to solve it.

    I just want to know if it is possible or not to get the volume of the given region if it is then how can i do it if it is not then i have to submit report to instructor that it is impossible to get this volume.
     
  11. Dec 3, 2009 #10

    Mark44

    Staff: Mentor

    Clarification on what exactly is the region to be revolved. As I said before, the description is poorly worded.

    If you want to proceed without getting clarification from you instructor, you can make an assumption about what you think the problem is saying, and then continue from there. I have laid out the steps for you. Have you read them? I'm somewhat surprised that this is post #10 in this thread and it doesn't seem that you have done anything in the way of working on this problem.


    Which given region? The problem description is unclear about the region to be revolved. That's what you should get clarification on. Once you know exactly what the region is, then you can find each volume of revolution using what I said in a previous post.
     
  12. Dec 3, 2009 #11
    I have already done the problem several times and i m not sure if i m right or wrong. This is a homework like i have to submit it.

    Y=sinx , y = cosx , x= pi/2 and the y-axis is the given region. This is giving me a square like shape ? i m confused how to break the volume of the region because two sin and cos curves are intersecting the the given region and also i have to consider the y axis.

    Please let me be clear how can i will break up the integral & the solution you gave is your made problem not mine & it is done with x-axis. Thats y is thread is getting bigger. Because i haven't found any solution yet.
     
  13. Dec 3, 2009 #12

    Mark44

    Staff: Mentor

    Well, I certainly can't tell if you're right or wrong because I haven't seen even the tiniest amount of your work.
    What I have repeatedly said is that the description you gave does not define an unambiguous region. There are three separate regions: one is below y = cos(x) and above y = sin(x) for x between 0 and pi/4, another is below y = sin(x) and above y = cos(x) for x between pi/4 and pi/2, and a third region that lies between the first two regions. Is the region we're suppose to revolve the first two and not the third or is it just the third, or what?
    What square like shape? What are you talking about? I don't see anything that looks remotely like a square.
    Why are you talking about breaking the volume of the region? The region - whatever it is - is two-dimensional, so has area but not volume. When we figure out what the region is, then we can revolve it around the y-axis (first problem) and get an integral that represents the volume. Same thing for the second problem, which entails revolving the region around the x-axis.
    This is not my made-up problem - I'm trying to understand what your problem is asking for. I have also said that when you have a problem you don't understand, you can either ask the instructor for clarification (have you done that?) or make some assumptions about what you think the problem is asking and go from there (have you done that?).
     
  14. Dec 4, 2009 #13
    These are the three regions i got. I guess i have to revolve them all.

    Sorry it is not a square like shape.
     
  15. Dec 4, 2009 #14

    Mark44

    Staff: Mentor

    You could ask your instructor what the region is ...
     
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