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Math Help - Volume/Convergence

  1. #1
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    Volume/Convergence

    Hello,

    A forewarning, this problem does take a lot of time. So thanks in advance for the help.

    The problem is as follows:

    "Let f(x) = [12(x^4 + 1)]/[2x^4 + 3x^2 + 2]. Consider the solid S formed by revolving the region between f(x) and the line y = 6 about the line y = 6.

    Decide whether the solid S has finite volume. Then, if it is finite, approximate it accurate to within .01. If it infinite, find a reasonable interval over which the volume is 100 or more."

    So, I started out by graphing it. I graphed f(x) and y=6. It appears there is a horizontal asymptote at y=6 as the graph goes out to infinity and negative infinity on each side. I'm used to doing problems around the x axis, using the p-test for when p > 1 it converges, and <= it diverges. Revolving it about y=6 also may change the result (as 1/x diverges, but when revolved around the x axis, converges).

    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by AfterShock
    Hello,

    A forewarning, this problem does take a lot of time. So thanks in advance for the help.

    The problem is as follows:

    "Let f(x) = [12(x^4 + 1)]/[2x^4 + 3x^2 + 2]. Consider the solid S formed by revolving the region between f(x) and the line y = 6 about the line y = 6.

    Decide whether the solid S has finite volume. Then, if it is finite, approximate it accurate to within .01. If it infinite, find a reasonable interval over which the volume is 100 or more."

    So, I started out by graphing it. I graphed f(x) and y=6. It appears there is a horizontal asymptote at y=6 as the graph goes out to infinity and negative infinity on each side. I'm used to doing problems around the x axis, using the p-test for when p > 1 it converges, and <= it diverges. Revolving it about y=6 also may change the result (as 1/x diverges, but when revolved around the x axis, converges).

    Thanks.
    Translate the function so that y=6 is moved so that it now coincides with the
    x-axis, that is consider the function:

    <br />
g(x)=f(x)-6<br />

    RonL
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  3. #3
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    Thanks Captain Jack. Are you able to help me solve the problem? It took your suggestion into consideration:

    int(f(x) - 6) [that is, just the integral; now I have to see whether rotating it around the x-axis will create a finite volue, as (1/x^2) does, or an infinite (1/x) with a finite area (Gabrielle's Trumpet)].

    So, I'm not sure of which test to go about doing this problem: integral, comparison, ratio, n-th term test, etc. The integral of [f(x) - 6] is:

    -9*sqrt(2)[(7*ln(2x^2 - sqrt(2)*x + 2) - 7ln(2x^2 + sqrt(2) + 2 + ... couple of arctangents, etc

    So, I don't think I want to use the integral test. Well, first off, I wouldn't want to use it on f(x) - 6; I would want to find cross-sections of the graph.

    Therefore: Pi*r^2; so in my case, taking the integral of Pi*[f(x) - 6]^2 will give me the volume, with the limits of integral being...0 to infinity? Or, negative infinity to infinity?

    Another question I have: since the integral test, ratio test, etc applies to only non-negative decreasing monotone functions (or so I thought), how would this apply to this problem, in order to find convergence. Maybe a comparison would be easier, but how would I apply that to finding the volume. I'm not sure. I'm quite lost right now.

    I think I have a lot of flaws in my thinking. But I needed to make some progress any way, so I tried giving it another shot.

    Thank you.
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  4. #4
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    Quote Originally Posted by AfterShock
    Thanks Captain Jack. Are you able to help me solve the problem? It took your suggestion into consideration:

    int(f(x) - 6) [that is, just the integral; now I have to see whether rotating it around the x-axis will create a finite volue, as (1/x^2) does, or an infinite (1/x) with a finite area (Gabrielle's Trumpet)].
    I suggest you consider:

    <br />
g(x)=f(x)-6<br />
,

    you are then interested in the integral of g(x).

    Now:

    <br />
g(x)=\frac{12(x^4+1)}{2x^4+3x^2+2}-6=\frac{-18x^2}{2x^4+3x^2+2}<br />

    So when |x| is large g(x)\sim-9/x^2, which
    should be sufficient to answer your question.

    RonL
    Last edited by CaptainBlack; May 8th 2006 at 10:48 PM. Reason: Error correction
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  5. #5
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    <br />
g(x)=\frac{12(x^4+1)}{2x^4+3x^2+2}-6=\frac{-18x^2-12}{2x^4+3x^2+2}<br />

    Where in the world did you get the -12 from? I agree they are the same, if you take out the "-12".

    So now I rotate Pi*g(x)^2 around the x-axis? What test do I use to determine convergence and such?
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  6. #6
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    Quote Originally Posted by AfterShock
    <br />
g(x)=\frac{12(x^4+1)}{2x^4+3x^2+2}-6=\frac{-18x^2-12}{2x^4+3x^2+2}<br />

    Where in the world did you get the -12 from? I agree they are the same, if you take out the "-12".
    An error - you should always check what you see here.

    So now I rotate Pi*g(x)^2 around the x-axis? What test do I use to determine convergence and such?
    I don't recall the names of these tests. It is clear that

    for |x|>N for large enough N that there exists a constant K>0
    such that:

    g(x)^2<K/x^4,

    and as \int_{-N}^N g(x)^2 dx is finite and \int_{N}^{\infty} K/x^4 dx and \int_{-\infty}^{-N} K/x^4 dx are both
    convergent the integral for the volume converges.

    RonL
    Last edited by CaptainBlack; May 10th 2006 at 12:02 AM.
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  7. #7
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    I've lost you on this part:

    |x| > N for large enough N that there exists a constant K > 0 such that,
    [g(x)^2]/[k/x^4]

    I understand the part after that, and I see how the comparison would work. I just think I need more algebra in there to convince me that that is indeed true. Knowing k/x^4 is larger and convergent, would then definitely mean g(x) converges, too. Assuming all of that is correct, I went on with approximating it accurate to within .01.

    Ok so we know g(x) <= int(1/x^4, x, 10, infinity). Now, we can make the integral to which we compared the original (and know is bigger) less than .01, therefore the original will be within .01, too.

    So, int(1/x^4, x, 10, infinity) = int(1/x^4, x, 10, N) + int(1/x^4, x, N, infinity). The latter is the tail.

    We want an N such that int(1/x^4, x, N, infinity) <= .01

    lim(as b--> infinity)[int(1/x^4), x, n, b] = -1/(3x^3)|(from n to b) which is equal to:

    0 - (-1/(3n^3)) = 1/(3n^3)

    So 1/(3n^3) <= .01 .... n <= 3.2183.... use n = 4.

    But I run into a problem... if that worked out correctly, wouldn't that be saying that:

    int(g(x), x, 10, 4) is within .01 ... I think I have basically got it but am missing minor detail.
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  8. #8
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    Disregard the second part; I did not square g(x) and I should have realized it didn't make sense because I tried to use a test for a non-negative integer series on an alternating series!

    Let me go through this a final time so I can get this problem done!

    g(x) = (-18x^2)/(24x^4 + 3x^2 + 2)
    g(x)^2 = (324x^4)/(2x^4 + 3x^2 + 2)^2

    So, now I have Pi * int((324x^4)/(2x^4 + 3x^2 + 2)^2, x, 1, infinity) <=
    int(k/x^4, x, 1, infinity) <= .01/Pi

    In which case, if I followed you correctly, why not just use k = 81, since that's what it is?

    Ok, so we're using int(81/x^4, 1, infinity) <= .01/Pi to make Pi*g(x)^2 also within .01. [Note: I am not sure if I am using the right limits of integration...can't use 0 since that's undefined].

    int(81/x^4, 1, infinity) = int(81/x^4, 1, N) + int(81/x^4, N, infinity). I will focus on the second part (the tail).

    int(81/x^4, N, infinity) <= .01/Pi

    Take the lim as b --> infinity from N to b, and we get,

    -27/x^3 evaluated from N to b is 0 - (-27/n^3) = 27/n^3 <= .01/Pi

    Solving for N I get: 20.3941...so I will use 21 for N. Therefore, if everything is right, I get this as a conclusion:

    The volume is finite, and int(g(x)^2, x, 1, 21) will approximate the volume to within .01.

    Anyone see any flaws?
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  9. #9
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    Quote Originally Posted by AfterShock

    Ok, so we're using int(81/x^4, 1, infinity) <= .01/Pi to make Pi*g(x)^2 also within .01. [Note: I am not sure if I am using the right limits of integration...can't use 0 since that's undefined].
    You want the area under the two tails to be <=0.01/pi as this bound the
    error in the truncated integral, so you are looking for an N such that:

    2*int(81/x^4, N, infinity) <= 0.01/pi

    int(81/x^4, 1, infinity) = int(81/x^4, 1, N) + int(81/x^4, N, infinity). I will focus on the second part (the tail).

    int(81/x^4, N, infinity) <= .01/Pi

    Take the lim as b --> infinity from N to b, and we get,

    -27/x^3 evaluated from N to b is 0 - (-27/n^3) = 27/n^3 <= .01/Pi

    Solving for N I get: 20.3941...so I will use 21 for N. Therefore, if everything is right, I get this as a conclusion:

    The volume is finite, and int(g(x)^2, x, 1, 21) will approximate the volume to within .01.

    Anyone see any flaws?
    Some flaws (as noted) but nice work and the right approach.

    If you have the facilities, you might want to check this numerically after
    you have recomputed N. (It might also be an idea to work with an error
    bound of 0.005/pi rather than 0.01/pi to allow for any possible problems when
    rounding to the nearest 0.01).

    RonL
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  10. #10
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    The infamous volume problem! I am doing something wrong, because when I check to see what the volume is by maple, I get conflicting results.

    First of all, how did you come up with this comparison:

    "I don't recall the names of these tests. It is clear that |X| > N for large enough N that there exists a constant K > 0
    such that:

    g(x)^2 < K/x^4"

    I am missing something, but how can I make it clear to someone that that is indeed true. Perhaps a little algebra?

    I am missing some link in my process.

    Let me briefly go through some of my thoughts again. And thanks, especially to CaptainBlack for helping with this problem!

    I'll try not to repeat the obvious steps.

    So, I have g(x)^2 = (324x^4)/(2x^4 + 3x^2 + 2)^2

    I am interested in finding the volume when revolving g(x) by the x-axis...it will be in the form Pi*int(g(x)^2, x = -infinity...infinity). I already calculated
    g(x)^2 as shown above. The answer MAPLE gets for the Volume is:

    [81*Pi^2*sqrt(14)]/49 ... which = 61.0454. So that's a reference point.

    Now, I am comparing it to Pi*int(81/x^4, x = -infinity...infinity). And we concluded that that is bigger than the integral we wanted to calculate. We want this new integral to be <= .01. However, we can divide our integral by Pi to get rid of it, and thus we want our error to be <= .01/Pi.

    I don't really get this part...you said I want the area of the TWO tails to be <= .01/Pi, so I want N such that
    2*int(81/x^4, x = -infinity...infinity) <= .01/Pi

    Would it be FOUR tails?

    This is where it gets tricky taking the integral, since my limits of integration are from -infinity...infinity. So since its symmetric along the y axis, just find from 0...infinity and then double it? So then I can see two tails only with which to deal.

    2*int(81/x^4, x = 1...infinity) =

    2*int(81/x^4, x = 1...N) + 2*int(81/x^4, x = N...infinity). I will focus on the tail, that is the last part (the from N to infinity one).

    So I take the lim as b --> infinity from N to b, and we get,

    -27/x^3 evaluated from N to b is 0 - (-27/N^3) = 27/N^3 <= .01/Pi

    But I have to remember to multiply by 2, so it's 2*[27/N^3] <= .01/Pi

    Solving for N, I get N = 25.695. Then, multiply by 2 again to get the volume from -infinity...1

    77.085.... that's not within .01...ahhhh im doing something wrong.

    Frustrating problem!
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