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Math Help - Integral

  1. #1
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    Integral

    I have the problem mostly done but this last part of the integration is getting to me. So we originally had a triple integral that I had to integrate with respect to z. The limits were from -(sqrt(13-4x^2))/3 and to the positive of that. So we ended up getting (z^3)/3 to plug in the limits to and, after simplifying, we have (2(13-4x^2)^3/2)/81 and I have to take that with respect to x. I simplified it a bit further and now am down to 26/81 on the outside to be multiplied in and (-8x^2 (sqrt(13-4x^2)) needing to be integrated. I was thinking substitution to get the sqrt to be u but then we would still be left with an x when we pull out the -8xdx. I was also thinking of doing it by part but I'm not sure if that would work either. Bah.
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    Quote Originally Posted by Centara View Post
    I have the problem mostly done but this last part of the integration is getting to me. So we originally had a triple integral that I had to integrate with respect to z. The limits were from -(sqrt(13-4x^2))/3 and to the positive of that. So we ended up getting (z^3)/3 to plug in the limits to and, after simplifying, we have (2(13-4x^2)^3/2)/81 and I have to take that with respect to x. I simplified it a bit further and now am down to 26/81 on the outside to be multiplied in and (-8x^2 (sqrt(13-4x^2)) needing to be integrated. I was thinking substitution to get the sqrt to be u but then we would still be left with an x when we pull out the -8xdx. I was also thinking of doing it by part but I'm not sure if that would work either. Bah.
    Let x = \frac{\sqrt{13}}{2}sin(\theta).

    -Dan
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  3. #3
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    Confused. Again. How did you get that and how would it help? This is why I'm dropping my math major once this class is over, lol. i can't see things like other people do.
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  4. #4
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    Hello,

    -(sqrt(13-4x^2))/3
    The trick is to get something like a-az, then to substitute z by sin(x) or cos(x).
    Hence \sqrt{a^2-(az)^2}=\sqrt{a^2(1-z^2)}=|a| \sqrt{1-sin^2(x)}=|a| \cdot |cos(x)|, which is easy to integrate. That's why we make this transformation

    ~~~~~~~~~~~~

    13-4x^2=(\sqrt{13})^2-\frac{13}{13} \cdot 2^2 x^2

    =(\sqrt{13})^2-(\sqrt{13})^2 \cdot \frac{2^2}{13} x^2

    a=\sqrt{13}

    \dots =a^2-a^2 \cdot \left(\sqrt{\frac{2^2}{13}} \cdot x \right)^2

    =a^2-a^2 \cdot \left(\frac{2}{\sqrt{13}} \cdot x \right)^2


    -----> \sin(\theta)=\frac{2}{\sqrt{13}} \cdot x

    ---------> x=\frac{\sqrt{13}}{2} \cdot \sin(\theta)

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  5. #5
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    Okay, so I honestly don't think my professor would give us something like that because that gives me a headache, lol. I'm wondering if maybe my limits are wrong or I messed up along the way. The question is: Find the volume bounded by the paraboloids y=9x^2 + 4z^2 and y=13-4x^2 - 9z^2

    I got the limits of y, the first of the equations being the lower and the second as the upper. Then the one's I found for z and then x being from -1 to 1. The equation i integrated was just 1 then dydzdx. Anyone see a problem with any of that? Or is it just a majorly hard problem?
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  6. #6
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    Quote Originally Posted by Centara View Post
    Okay, so I honestly don't think my professor would give us something like that because that gives me a headache, lol. I'm wondering if maybe my limits are wrong or I messed up along the way. The question is: Find the volume bounded by the paraboloids y=9x^2 + 4z^2 and y=13-4x^2 - 9z^2

    I got the limits of y, the first of the equations being the lower and the second as the upper. Then the one's I found for z and then x being from -1 to 1. The equation i integrated was just 1 then dydzdx. Anyone see a problem with any of that? Or is it just a majorly hard problem?
    The paraboloids intersect on the circle x^2 + z^2 = 1. So you have to integrate (13 - 4x^2 - 9z^2) - (9x^2 + 4z^2) in the xz-plane over the region bounded by this circle. Switching to polar coordinates would probably be a smart thing to do.
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