1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook