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Homework Help: Find the volume of the region

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the region bounded by y = x^2 and the line y = 1. Find the exact volume of the solid obtained by rotating the region about the x-axis.

    2. Relevant equations

    3. The attempt at a solution

    ∫ π(1-x^2)^2dx from 0 to 1

    (πx^5)/5 -(π2x^3)/3 +πx evaluated from 0 to 1

    π/5 - π2/3 +π = (3π-10π+15π)=8π/15

    This answer is wrong. What am I doing wrong?
  2. jcsd
  3. Dec 7, 2012 #2


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    x=0 isn't an x value where the two curves intersect. x=1 is. There's another one.
  4. Dec 7, 2012 #3


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    The lower limit is wrong. Graph the region.

    The integrand is wrong. Check your formula for the disk or washer method.

  5. Dec 7, 2012 #4
    i got my answer to be 4π/5

    my integral was setup as 2pi ∫(x^2)^2 dx from -1 to 1

    I'm not sure how I came about this answer.

    If I use the washer method I end up with ∫ π(0)^2 - π(x^2)^2 from -1 to 1

    ∫ -π(x^4) dx from -1 to 1 the answer to this is the negative half of what I got. I did several other similar examples and they all displayed the same pattern. I'm still not sure what I'm doing wrong :/
  6. Dec 7, 2012 #5


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    Are you trying to figure out what's going on with examples in your notes? If so, why don't you look away from them and just try this one from scratch. It kind of looks like you are trying to find the volume of y=x^2 rotated around the x-axis. That would be the 'negative half' that you subtract from the volume of the cylinder you get from rotating y=1 around the x-axis. Is that it?
  7. Dec 7, 2012 #6
    I'm not using any notes. The examples I'm getting are randomly generated online. I'm trying to get the volume of the solid in between y=1 and y=x^2 rotated around the x-axis. I'm guessing it should look something like this?:confused:


    As for creating an actual expression I'm not sure how to approach this problem.

    I guess the interval would be from -1 to 1 since y=1 and 1=x^2 x=+-1 and x=0 at x=0 y=0

    I'll try again I suppose.
    ∫ pi(1^2)-pi(x^2)^2 dx

    that ends up giving me 8pi/5. The actual answer is 4pi/5 which i can get if the interval is from 0 to 1. My guess is that the problem was improperly worded and forgot to mention x=0 as one of the bounds. Either way thanks for your help. I really appreciate it!
    Last edited: Dec 7, 2012
  8. Dec 8, 2012 #7


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    That all seems ok to me.
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