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Math Help - Showing that the differential form is exact;

  1. #1
    s7b
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    Showing that the differential form is exact;

    I started this question and got stuck;

    Show that the differential form in the integral is exact. Then evaluate the integral:

    Integral from (0,0,0) to (3,3,1) of 2xdx - y^2dy - 4/(1+z^2)dz

    This is what I've done so far:

    M=2x , N=-y^2 , P=-4/(1+z^2)

    Then I showed that it is conservative because;
    dp/dy=dn/dz , dm/dz=dp/dx , dn/dx=dm/dy
    *(I wasn't sure what it meant by showing it was exact, I thought that meant conservative)*

    from here I did;

    df/dx=2x ---> f(x,y,z)=x^2 + g(y,z)

    But then df/dy=x^2 + dg/dy = y^2 so this is where I didn't understand what to do....

    Please help if you know this!!!
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    Quote Originally Posted by s7b View Post
    I started this question and got stuck;

    Show that the differential form in the integral is exact. Then evaluate the integral:

    Integral from (0,0,0) to (3,3,1) of 2xdx - y^2dy - 4/(1+z^2)dz

    This is what I've done so far:

    M=2x , N=-y^2 , P=-4/(1+z^2)

    Then I showed that it is conservative because;
    dp/dy=dn/dz , dm/dz=dp/dx , dn/dx=dm/dy
    *(I wasn't sure what it meant by showing it was exact, I thought that meant conservative)*

    from here I did;

    df/dx=2x ---> f(x,y,z)=x^2 + g(y,z)

    But then df/dy=x^2 + dg/dy = y^2 so this is where I didn't understand what to do....

    Please help if you know this!!!

    You are correct in that f(x,y,z)=x^2+g(y,z) but then

    \frac{\partial f}{\partial y}=\frac{\partial g}{\partial y}=-y^2 \implies g(y,z)=-\frac{1}{3}y^3

    So now f(x,y,z)=x^2-\frac{1}{3}y^3+h(z)

    \frac{\partial f}{\partial z}=\frac{\partial h}{\partial z}=-\frac{4}{1+z^2} \implies h(z)=-4\tan^{-1}(z)

    Now f(x,y,z)=x^2-\frac{1}{3}y^3-4\tan^{-1}(z)

    We can show it is exact becuase

    \nabla f= 2x \vec i -y^2 \vec j -\frac{4}{1+z^2}\vec k

    This does show that is is a conservative vector field so the fundemental theorem of line integrals applys and

    so the value is

    f(3,3,1)-f(0,0,0)=(3^2-\frac{1}{3}3^3-4\tan^{-1}(1))-(0^2-\frac{1}{3}0^3-4\tan^{-1}(0))=-\pi
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