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Math Help - Triple Integral

  1. #1
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    Triple Integral

    \int\int\int xy dV where 0 <= x <= 3 & 0 <= y <= x & 0 <= z <= x+y

    Thought it'd be simple, but apparently I'm missing something because my answer isn't matching the back of the book.

    Help's appreciated.
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  2. #2
    Moo
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    Hello,

    The bounds for z contain x and y, so you'll have to integrate firstly in respect with z.

    Then, the bounds for y contain x, so integrate in respect with y before x.


    \int_0^3 \int_0^x \int_0^{x+y} xy \ dz dy dx

    \int_0^3 \int_0^x \left[xyz\right]_0^{x+y} dy dx

    \int_0^3 \int_0^x xy(x+y) \ dy dx

    \int_0^3 \int_0^x yx^2 \ dy dx + \int_0^3 \int_0^x xy^2 \ dy dx

    \int_0^3 x^2 \left[\frac{y^2}{2}\right]_0^x dx + \int_0^3 x \left[\frac{y^3}{3}\right]_0^x dx

    \int_0^3 x^2 \frac{x^2}{2} dx + \int_0^3 x \frac{x^3}{3} dx

    \frac{1}{2} \int_0^3 x^4 dx + \frac 13 \int_0^3 x^4 dx

    \frac 56 \int_0^3 x^4 dx

    \frac 56 \left[\frac{x^5}{5}\right]_0^3

    \frac 56 \times \frac{3^5}{5}

    \frac{81}{2}


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  3. #3
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    Hmhm, you got the same answer I did...glad I'm not alone, at least. ^^

    The answer in the back of the book is 64/15.
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