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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.

    Lol brb, withdrawing my UCAS application.
    OP that almost made me sick.

    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


    \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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