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Homework Help: Solids of revolution question !

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the volumes of the solids generated

    a) x-3y+3= 0, x=0 , y=2 about the x axis

    I sketched the graph got a straight line,

    I then proceeded to integrate y2

    ∏∫y2δx

    =[(x^3/27)+(x^2/3)+ (x)] from x=3 to x=-3
    I got 8∏
    but the answer is 5∏


    b) x-y2-1=0, x=2 about the y-axis

    ∏∫x2δy

    = [ (y^5/5) + (2y3/3) + (y)] from x=1 to x=-1

    I got (56/15)∏ for this one but the answer is (64/15)∏
     
  2. jcsd
  3. Feb 20, 2013 #2
    a) First off, don't memorize these formulas, you have to understand what's going on.
    You're asked to find the area between:
    y=(x+3)/3, x=0, y=2 about the x-axis
    So the graph looks like this:
    attachment.php?attachmentid=55944&stc=1&d=1361423402.png

    Now, the way to do this problem is that you need to subtract the outer radius from the inner right?

    So what does the integral become?
     

    Attached Files:

  4. Feb 21, 2013 #3
    Uh is it the intergral of (x/3) +1 - (integral of y=2)?
     
  5. Feb 21, 2013 #4

    HallsofIvy

    User Avatar
    Science Advisor

    First, it is clear from the graph that the line y= 2 is above the line y= x/3+ 1. Your calculation will give a negative volume!

    Second, you want to subtract the volumes, not just the integrals! The volume given by y(x) rotated around the x-axis is [itex]\pi\int y^2 dx[/itex].
     
  6. Feb 21, 2013 #5
    Adding onto what HallsofIvy said, what you're doing is subtracting the volumes of the two rotations, so technically you can have 2 integrals but they would be:
    [tex]\pi\int (outer)^{2}dx-\pi\int (inner)^{2}dx[/tex]
    This is just using pi r^2 to find the area of a piece of the circle, the dx would be the width of the circle and the integral because it is the sum of many pieces between 2 points.
    Do you see why that works?
     
  7. Feb 21, 2013 #6
    ∏∫(4)^2dx− ∏∫(((x/3)+1)^2dx ?

    Oh when I do this from x=0 to x=3 I do get 5∏
     
  8. Feb 21, 2013 #7
    I got the 2nd one too. Thanks everyone
     
  9. Feb 21, 2013 #8
    That's correct.
     
  10. Feb 21, 2013 #9
    Oh may I ask one more question? I think I got the answer but not sure if my method was correct.
     
  11. Feb 21, 2013 #10

    Mark44

    Staff: Mentor

    Which problem are you asking about? If it's the first problem, you can do this in a couple of ways: by "washers" (disks with holes in them) or by cylindrical shells (sort of like the layers in an onion).

    Either way should produce the correct answer, but often one way will be easier than the other, so it pays to be comfortable using both ways. Also, the advice given by iRaid to not memorize the integration formulas is good advice. However, you do need to know the formulas for the area of a circle and for the area of a rectangle, both of which are pretty simple.

    To get the volume of a disk, it's just the area of the circle times the thickness of the disk, which will usually be either Δx or Δy.
    To get the volume of a washer, find the volume of the outer disk, and subtract the volume of the inner disk (the hole).
    To get the volume of a shell, you need the area of a rectangle times its thickness. One of the dimensions of the rectangle comes from the circumference of the cylindrical shell, which would be ##2\pi## times the radius of the shell. The length is just the length of the shell, and the thickness is either Δx or Δy.

    It is extremely important to draw a picture of the region that will be rotated, and as good a sketch of the solid of revolution as you can manage. If you don't draw these pictures, there's a very good chance that the integral you set up will not produce the correct answer.
     
  12. Feb 21, 2013 #11
    Well the question was to find the volume generated by

    y=1/x , y=1 , x=2 about y=0

    I did this

    ∏∫(1/x^2).dx from x=0 to x=2 I got -1/2∏

    then ∏∫1.dx =[x] from x=0 to x=2 and I got ∏

    then I did this ∏-(1/2)∏ = 1/2∏

    is this correct?
     
  13. Feb 21, 2013 #12

    Mark44

    Staff: Mentor

    Do you mean between x = 1 and x = 2?

    Also, is the region between the graph of y = 1/x and the x-axis?

    Try to be more precise in what you ask.
     
  14. Feb 21, 2013 #13
    Find the volumes of the solids generated when each of the areas enclosed by the following curves and lines is rotated about y=0

    y=1/x , y=1 , x=2
     
  15. Feb 21, 2013 #14

    Mark44

    Staff: Mentor

    One of you limits of integration is wrong. Did you sketch the region that is being rotated?
     
  16. Feb 21, 2013 #15
    Uhh I tried I sketched y=1/x and drew in the line y=1

    and the region is somewhere between y=1/x and y=1 right?
     
  17. Feb 21, 2013 #16

    Mark44

    Staff: Mentor

    Right, and the region is also bounded by the line x = 2. So where does the curve intersect the line y = 1?
     
  18. Feb 21, 2013 #17
    The curve meets y=1 at x=1
     
  19. Feb 21, 2013 #18

    Mark44

    Staff: Mentor

    Yes, so you should be integrating from 1 to 2, not from 0 to 2.
     
  20. Feb 21, 2013 #19
    Oh... so that's how I should get 1/2pi thanks
     
  21. Feb 21, 2013 #20

    Mark44

    Staff: Mentor

    That's not what I'm getting. How did you get that?
     
  22. Feb 22, 2013 #21
    lionely, you have to understand the concept of this. Believe me, if you just learn the formulas and not HOW this stuff works, you'll fail during the exam. The next thing you're going to be learning is the shell method which builds upon this even more. When your teacher starts throwing in rotations around lines that aren't the x and y axis, you'll be completely screwed.
     
  23. Feb 22, 2013 #22
    But I thought I should integrate 1/x^2

    then evaluate [-1/x] from x= 1 to x =2?
     
  24. Feb 22, 2013 #23

    Mark44

    Staff: Mentor

    That's not the whole integrand.
    These are the right integration limits, but it looks like you don't have the right integrand.

    I don't think you have a drawing (or maybe a good drawing) of the region that's being rotated. The region being rotated is the roughly triangular region above the curve y = 1/x, below the line y = 1, and to the left of the line x = 2.

    Each area element is a vertical strip of width Δx. Each area element, when rotated, is in the shape of a washer whose volume is the area of the washer times its width, Δx.
     
  25. Feb 22, 2013 #24
    OKAY SIR I THINK I got it if not I truly hate integration , xD.

    I got 1/2pi from evaluating [-1/x]

    NOW I now find the volume under y=1 for the same interval
    I get pi NOW
    I then subtract the volumes ( Pi - 1/2pi) = 1/2pi cu. units.
     
  26. Feb 22, 2013 #25

    Mark44

    Staff: Mentor

    Edit: I made a mistake in evaluating the antiderivative.

    [STRIKE]No, that's the same as you got earlier, and it's still not right.
    [/STRIKE]
    What does your integral look like?

    $$ \pi \int_1^2 (f(x)) dx$$

    If you "Quote" my post, you can use my LaTeX above and fill in the integrand.
     
    Last edited: Feb 22, 2013
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