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Math Help - Trouble with volume/area problems with integrals

  1. #1
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    Trouble with volume/area problems with integrals

    Okay, so I'm almost finished my webwork assignment, but 3 problems still evade me. I'll say what they are and show what I have done so far.

    1. FInd the area of the region enclosed by the curves 2y=3sqrt(x), y=5 and 2y+4x=7.

    2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=3+(1/x^2), y=3, x=4 and x=8 about the x-axis.

    So for 1, I drew a sketch based on what it looked like on my graphing calculator. Nothing too fancy, but just enough so I would know to divide my almost triangular shape into two intervals.

    I found the intercepts to be (going from left to right), x=-3/4, 1 and 100/9. I found -3/4 by substituting 5 for y into 2y+4x=7. I found x=1 by setting 3sqrt(x)=7-4x, squaring both sides and solving the quadratic, but I suspect that this may be my mistake. And I found x=100/9 by substituting y=5 into y=(3sqrt(x))/2.

    I did it in two parts.
    Afinal=A1+a2

    A1= integral(-3/4 to 1) (5-((-4x+7)/2))dx

    I eventually got 0.8125 for this part.

    A2=integral (1 to 100/9) (5-((3sqrt(x))/2)

    I got 18.51851852 for this part.

    When added, I got a final answer of 19.33101852.

    I have no idea what I'm doing wrong, but I'm sure it must be something really little and annoying.
    Last edited by madvie; July 18th 2008 at 01:16 PM.
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  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
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    Taguig City, Philippines
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    why dont you do it horizontally?

    solve for x in terms of y..

    (1) x=\frac{4}{9}y^2
    (2) x = \frac{7-2y}{4}

    get the intersection of that 2 functions, and that would be (1,3/2)

    then, \int_{3/2}^5 \left(\frac{4}{9}y^2 - \frac{7-2y}{4}\right) \, dy
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