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Math Help - Find the volume of the solid.

  1. #1
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    Find the volume of the solid.

    I'm stuck on this problem. The answer is probably obvious, but I second guess myself a lot.

    Find the volume of the solid below z=sqrt((x-y)(x+4y)), above the parallelogram in the x-y plane with vertices (0,0)(1,1)(5,0)(4,-1).

    I have drawn myself a graph to visualize.
    Since I havent figured out the way to input stuff on this forum yet, bare with me.

    My setup so far is:
    triple integral : integral from 0 to 5, integral from -1 to 1, integral from 0 to Z eqn above dzdydx.
    The hint given with the problem is change of variables. I've started working it, but since I cant recall what change of variables Im looking to use, my work doesnt seem right so far.
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  2. #2
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    Are you familiar with jacobians? You should integrate using a double integral of the form:

    \int_R\int f(x,y)dxdy=\int_S\int f(g(u,v),h(u,v)))\mid\frac{\partial (x,y)}{\partial (u,v)}\mid dudv

    Find the equations of the lines of the parallelogram so that you can find a transformation T from S in (u,v) to R (x,y) such that S is rectangular. Look in your book for some guidance, I'll help you find the region if you get stuck.

    The transformation is a function T(u,v)=(g(u,v),h(u,v))

    EDIT: Sorry, I worded that wrong and fixed it. I orignally mentioned a transformation T from S in (x,y) to R, however S is in the (u,v) plane NOT (x,y)... I hope I didn't cause any confusion...
    Last edited by adkinsjr; December 12th 2010 at 10:52 PM.
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  3. #3
    opt
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    Hi dmbocci,
    First, answer the question: what is the domain that you are integrating over?
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  4. #4
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    I know of the jacobian, although we didnt spend very much time on it. So basically I'm figuring out this one from scratch. I have an idea though. I have the equations of the lines of the parallelogram. I have y=x and y=(-1/4)x.
    So u=x-y and v=-1/4x-y..? Then it gets a bit confusing..
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  5. #5
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    Here are the lines:

    y=-\frac{1}{4}x+\frac{5}{4}
    y=-\frac{1}{4}
    y=x
    y=x-5

    Then you can rewrite them:

    4y+x=5
    4y-x=0
    y-x=0
    x-y=5

    Now if you let u=y-x and v=4y+x the lines u=0 and v=5 are going to be two of the four lines that bound your region in the u-v plane. So you're going to have a simple rectangular region. Solve these equations for x and y to obtain functions x=g(u,v) and y=h(u,v) by solving the u and v equations for x and y, I obtained:

    x=\frac{1}{5}(v-4u)

    y=\frac{1}{5}(u+v)

    Now your region S should be -5\leq u\leq 0, 0 \leq v\leq 5

    Does this make sense so far?
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  6. #6
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    Quote Originally Posted by dmbocci View Post
    I know of the jacobian, although we didnt spend very much time on it. So basically I'm figuring out this one from scratch. I have an idea though. I have the equations of the lines of the parallelogram. I have y=x and y=(-1/4)x.
    So u=x-y and v=-1/4x-y..? Then it gets a bit confusing..
    This would work, there are multiple possibilities. As long as S turns out to be reasonable.
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  7. #7
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    The integral I obtained looks like this:

    \int_{-5}^0\int_0^5\sqrt{\frac{1}{5}(-uv)}\mid\frac{\partial (x,y)}{\partial (u,v)}\mid dvdu

    This should work, don't worry about the negative under the square root, this can be taken as \sqrt{-\frac{1}{5}u}\sqrt{v} and the function \sqrt{-\frac{1}{5}u} is defined on [-5,0]...

    The Jacobian will be a number, I didn't find it, that should be easy enough.

    A solution can be found here as well:

    Calc Chat Free Solutions

    9th edition, 14.8 problem 25...
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  8. #8
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    oh ok, I think I see it for the most part. Thank you very much!
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