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Math Help - Double Integral

  1. #1
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    Double Integral

    Find the volume of the solid lying under the plane z=4x+3y+5 and above the rectangle R={(x,y) | -1<= x <= 0 , 2 <=y <=3}

    \int_{2}^{3}\int_{-1}^{0}~(4x+3y+5) dxdy

    =~\int_{2}^{3}~[2x^2+3yx+5x]_{-1}^{0}~dy

    =~\int_{2}^{3}~(-\frac{3y}{2} - \frac{11}{6})dy

    = ~ [-\frac{3y^2}{4} - \frac{11y}{6}]_{2}^{3}

    =~-\frac{28}{3}

    which is wrong. The correct answer is \frac{21}{2}

    Can someone help me? I have a feeling I might have miss used -1,0 and 2,3 maybe.
    I also have another question similar to this I will need help on because I'm not quite sure where to start.
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  2. #2
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    There's nothing wrong with your set up, must be in the algebra.

    I get 21/2.
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  3. #3
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    on the last step I even put it in the calculator and get -28/3. Can't seem to figure out where I have messed up.
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  4. #4
    Eater of Worlds
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    I don't know what you're doing. I am getting 21/2. I even tried various arrangements and did not arrive at -28/3
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  5. #5
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    My other question was I need help on this problem as I don't know where to even start. Thanks

    Find volume of the solid bounded by the surface
    z=x\sqrt{9x^2+y}
    and the plane x=1,x=0,y=1,y=0 and z=0.
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  6. #6
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    Quote Originally Posted by dukebdx12 View Post
    =~\int_{2}^{3}~[2x^2+3yx+5x]_{-1}^{0}~dy

    =~\int_{2}^{3}~(-\frac{3y}{2} - \frac{11}{6})dy
    Check this. (It's just a matter of arithmetic.)
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    Check this. (It's just a matter of arithmetic.)
    got it now...thankyou
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