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Solid of Revolution of an arcsin graph

  1. Oct 23, 2011 #1
    1. Consider the function f(x) = arcsin(2x^2 - 1).
    Write down, but do not attempt to solve, a definite integral in terms of y, which when evaluated will give the volume of the solid of revolution formed by rotating the graph about the y-axis.




    2. Endpoints are at (-1, pi/2) and (1, pi/2).
    x-intercepts are at (-1/sqrt(2), 0) and (1/sqrt(2), 0).
    y-intercept is at (0, -pi/2).
    The equation looks like a "bowl" shape.




    3. I know that there are two main methods: the disc method and the shell method. Unfortunately, I don't know how to apply them very well. I'm used to having functions only in one quadrant, but this function is in all 4 quadrants at once!
    My best attempt would be something like this:
    Trying to put it in the form: 2pi*r*h
    Integral from -1 to 1 of: 2pi * arcsin(2x^2 - 1) * dy
     
  2. jcsd
  3. Oct 23, 2011 #2

    HallsofIvy

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    You need of course, to write that integrand in terms of the variable of integration, y.

    If y= arcsin(2x^2- 1, then 2x^2- 1= sin(y), 2x^2= 1+ sin(y), x= sqrt{(1+ sin(y)/2). By symmetry, it is sufficent to find the volume generated by rotating the graph in the first quadrant around the y- axis and then multiply by two.
     
  4. Oct 30, 2011 #3
    So once I have it in the form x = sqrt((sin(y) + 1)/2), do I then have to change the numbers on the top and bottom of the integral? So instead of going from -1 to 1, I am now going from -pi/2 to +pi/2? And by doing that, I am now wrapping it around the x-axis? So I can change it to become:
    y = sqrt((sin(x) + 1)/2) from -pi/2 to pi/2 around the x-axis?
    If that is correct, what do I do from then on to create the definite integral?
     
  5. Oct 30, 2011 #4
    You said I should cut it in half and then calculate half of that. Does that mean I am only going from 0 to +pi/2 or from -pi/2 to 0, rather than the full way?
    So would the integral be like this:
    Integral from 0 to +pi/2 of: (sqrt((sin(x) + 1)/2)) * dx
    Is that the answer?
     
  6. Nov 5, 2011 #5
    Please someone help me! It's been literally weeks and I still can't solve this problem :(
     
  7. Nov 7, 2011 #6
    Okay, my new attempt at the solution is to use the formula:
    V = Integral from a to b of [f(y)]^2 * dy
    By transposition we get:
    x = sqrt((sin(y) + 1) / 2)
    Since x = f(y), then [f(y)]^2 = (sin(y) + 1) / 2
    Now I thought that since an arcsin graph is symmetrical, I can just simply calculate it from -pi/2 to pi/2 since when it rotates it's going to cover the other half anyway.
    So the definite integral goes from -pi/2 to pi/2
    V = Integral from -pi/2 to pi/2 of (sin(y) + 1) / 2 * dy
    Is that correct?
     
  8. Nov 13, 2011 #7
    Please tell me if I'm doing it right or wrong because I am still confused!
     
  9. Nov 16, 2011 #8
    It's been almost a month and I still haven't solved it! Please help me!
     
  10. Nov 16, 2011 #9
    I think what happens is people see a list of answers and just assume it's answered. Just use volume by shells along the y-axis where the volume of the i'th shell is just [itex]\pi r^2 dy[/itex] right? but keep in mind, wish to have the radius always positive so when you solve for the radius in terms of x, you get:

    [tex]x^2=1/2 \sin(y)+1[/tex]

    but y goes from -pi/2 to pi/2 and the negative values would give a negative value of the sine. So once you get that straight, you can write the volume as:

    [tex]V=\pi \int_{-\pi/2}^0 (1/2\sin(-y)+1)dy+\pi \int_0^{\pi/2}(1/2\sin(y)+1)dy[/tex]

    think so anyway. You go through it and make sure everything makes sense to you.
     
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