Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 1

Put $r=\sqrt{x^2+y^2}$. The definitions of the parabolic polar coordinates give $u^2=x+r$ and $v^2=-x+r$. Subtracting and multiplying these equations yieldsSince $u,v,y\ge 0$, taking the nonnegative square root givesThe Jacobian matrix has columns $(u,v)$ and $(-v,u)$. Its Jacobian determinant is thereforeHere the displayed fraction denotes the Jacobian determinant, as in the paper. The change of variables formula consequently gives $dxdy=(u^2+v^2)dudv$. The Jacobian determinant is positive except at the origin, and the definitions supply the unique nonnegative inverse throughout the region $y\ge 0$.

For $u=1$, the inverse parabolic polar coordinates give $y=v$ and $x=(1-v^2)/2$. For $v=1$, they give $y=u$ and $x=(u^2-1)/2$. Thus the requested curves are the upper halves of the parabolasThe first parabola has vertex $(1/2,0)$ and opens to the left; the second parabola has vertex $(-1/2,0)$ and opens to the right. Both pass through $(0,1)$ and are reflections of one another across the $y$-axis.

The inequalities become the rectangle $1/2\le u\le 1$, $1/2\le v\le 1$ in parabolic polar coordinates. The change of variables formula and the Jacobian determinant from a give the areaThe middle panel in b shows the curved region in Cartesian coordinates; its straight edges in parabolic polar coordinates explain the simple integration limits.

The first boundary is $u=1$. On $y=0$ in the first quadrant, the inverse parabolic polar coordinates give $v=0$. To identify the remaining curve, put $t=y^{1/3}\ge 0$. Its Cartesian coordinates satisfyHence $u=t$ and $v=t^2$, so this boundary is $v=u^2$. The condition $x\ge 0$ on this curve requires $0\le t\le 1$. Its endpoints are $(0,0)$ and $(0,1)$, while $u=1$ meets $v=0$ at $(1/2,0)$. These three arcs enclose exactlyEvery point in this region has $v\le u$, and therefore $x=(u^2-v^2)/2\ge 0$, as required. Using the Jacobian determinant, the area is