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    Here is the question I have been given and as I have no resources to check my answers I am posting it here for any help possible.

    Here is my answer:

    i) Don't know how to do this on here lol

    ii) S is the region bounded by the parabolas y^2={\alpha}x, y^2={\beta}x, x^2={\gamma}y, x^2={\delta}y

    0<\alpha<\beta \mathrm{&}, 0<\gamma<\delta

     u=\frac{y^2}{x}, v=\frac{x^2}{y}, u_x=\frac{-y^2}{x^2}, u_y=\frac{2y}{x}, v_x=\frac{2x}{y}, v_y=\frac{-x^2}{y^2}

    So the Jacobian is: \begin{pmatrix} \frac{-y^2}{x^2} & \frac{2y}{x} \\\frac{2x}{y} & \frac{-x^2}{y^2} \end{pmatrix}

    iii) I=\int\int_S \frac{1}{xy} dxdy, u.v=\frac{x^2y^2}{xy}=xy \Longrightarrow \frac{1}{xy}= \frac{1}{uv}

    I= \int\int_S \frac{1}{uv}dudv=\int \frac{ln(u)}{v}=ln(u)ln(v)

    u= \frac{y^2}{x} v= \frac{x^2}{y} \Longrightarrow I=ln(\frac{y^2}{x})ln(\frac{x^2}  {y}

    Are my answers correct or should I be doing something different? Thank you.
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    Lol brb, withdrawing my UCAS application.
    OP that almost made me sick.
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    You haven't used the determinant of the Jacobian in the change of variable in iii).

    \displaystyle\int \int_S \dfrac{1}{xy} \ \ dx dy = \displaystyle\int \int_S \dfrac{1}{uv} \left\lvert \dfrac{\partial (u,v)}{\partial (x,y)} \right\rvert \ \ du dv

    You should notice, your change of variable in iii) is actually completely meaningless, you've gone from

    \displaystyle\int \int_S \dfrac{1}{xy} \ \ dx dy

    to

    \displaystyle\int \int_S \dfrac{1}{uv} \ \ du dv

    i.e. all you've done is a relabelling

    Also, you need to use the boundary conditions of the region S to actually evaluate the integral.
 
 
 
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