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Let us consider the mapping $\phi:\mathbb{R}\to\mathbb{R}$ given $\phi(u,v):\begin{cases} x=u+v\\y=v-u^2 \end{cases}$

Let $D$ denote the triangle with vertices $(0,0),\,(2,0)$ and $(0,2)$ in the plane $(u,v)$. Show that $\phi$ is a change of variables in a neighborhood of $D$. Find the set $\phi(D)$ and calculate its area.

I have taken the values of $u$ and $v$ from the coordinates of the vertexes given, and I have used them to calculate corresponding coordinates for $x$ and $y$, namely $(0, 0),\, (2, -4)$ and $(2,2)$

I have also calculated the equation for the hypotenuse of the triangle $D$, that is $v = 2 - u$. I tried determining values for $x$ and $y$ form this, and got $x = 2$ and $y = (1-u)(u+2)$.

From this point I'm really not sure how to proceed. I am finding it confusing working backwards to find the original region when given the transformation.

Many thanks for any help given!

John Wayland Bales
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Heav
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2 Answers2

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Since the triangle $D$ is right-angled with legs on the $u$-axis and $v$-axis, the first thing to do is to determine how these axes map to the $(x,y)$ coordinate system. We observe that the line $v = 0$ (i.e. the $u$-axis) becomes $(x,y) = (u, -u^2)$: this is a parametric equation for a parabola with vertex at $(x,y) = (0,0)$, and axis of symmetry $x = 0$. So we know that one leg of $D$ maps to a parabolic arc: specifically, since $u \in [0,2]$ for this leg, we find that this parabolic arc is located in quadrant IV, and has endpoints $(0,0)$ and $(2,-4)$.

For the line $u = 0$ (i.e. the $v$-axis), we have $(x,y) = (v,v)$: this is the line $y = x$. For $v \in [0,2]$, this is just the line segment joining $(x,y) = (0,0)$ and $(x,y) = (2,2)$.

The hypotenuse of $D$ is trickier, but a simple way to do it is to note it is a segment of the line $v = 2-u$; thus we can find a parametrization with respect to $u$ in the $(x,y)$ plane: $$(x,y) = (u+(2-u), (2-u)-u^2) = (2, 2-u-u^2).$$ This is a vertical line at $x = 2$. And of course, this line must join the aforementioned segments at the vertices of $D$ mapped under $\phi$.

From here, it is not difficult to sketch and describe the region $\phi(D)$.

heropup
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There are three sides of the triangle in the $(u,v)$ domain.

  1. Side $(u,v)=(0,0)$ to $(u,v)=(2,0)$. So $v=0$ which means $x=u,\,y=-u^2$ therefore in the $(x,y)$ domain this side of the triangle will be mapped into the parabola $y=-x^2$. The point $(u,v)=(0,0)$ goes to the point $(x,y)=(0,0)$ and the point $(u,v)=(2,0)$ goes to the point $(x,y)=(2,-4)$, so that is the segment of the parabola to which that side of triangle is mapped.

  2. Side $(u,v)=(0,0)$ to $(u,v)=(0,2)$. So $u=0$ which means that $x=v$ and $y=v$ therefore in $(x,y)$ domain this side of the triangle will be mapped into the line $y=x$. The point $(u,v)=(0,0)$ goes to the point $(x,y)=(0,0)$ and the point $(u,v)=(0,2)$ goes to the point $(x,y)=(2,2)$, so that is the segment of the line to which that side of triangle is mapped.

  3. Side $(u,v)=(2,0)$ to $(u,v)=(0,2)$. This line has equation $u+v=2$ which translates directly to $x=2$, so this side of the triangle is mapped into the line $x=2$, specifically along the segment from $(x,y)=(2,-4)$ to $(x,y)=(2,2)$.

The diagram below shows both the original triangle $(u,v)$ domain and the $(x,y)$ domain using two different hues.

Domains of integration

John Wayland Bales
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