folks.
I've got this question:
Let $D$ be the region $\{(x,y) ~|~ 0 \leq y \leq x, 0 \leq x \leq 1\}$. Evaluate:
$$\iint_D (x + y) dxdy$$ by making the change of variables $x = u + v$, $y = u – v$. Check your answer by evaluating the integral directly by using an iterated integral.
I've done the problem a couple of different ways, and have gotten a couple of different answers that are kind of the correct answer but aren't the correct answer (which is $\displaystyle\frac{1}{2}$). Neither of the methods I've tried have worked, and the fact that they're all giving me different answers is indicative of some intrinsic mishap that I'm just not catching.
I take the Jacobian determinant of $f(x,y) = (u + v, u – v)$, which is $-2$. I solve for $u$ and $v$ in terms of $x$ and $y$ to get a transformation, $\displaystyle T(x,y) = \left ( \frac{x + y}{2}, \frac{x – y}{2} \right )$, to use on the boundary points of $D$.
If we draw $D$ out, we get a triangle whose vertices are the points $(0,0), (1,0),$ and $(0,1)$. These boundary lines can be parameterized by the functions:
$$\left.
\begin{array}{1}
C_1(t) = (t, 0) \\
C_2(t) = (t, t) \\
C_3(t) = (1,t)
\end{array}
\right \} ~ ~ ~ ~ \text{for} ~ ~ ~ ~ 0 \leq t \leq 1$$
The transformation of these boundary points is, then:
$$\left.
\begin{array}{1}
T(C_1(t)) = \left (\frac{t + 0}{2}, \frac{t – 0}{2} \right) \\
T(C_2(t)) = \left (\frac{t + t}{2}, \frac{t – t}{2} \right) \\
T(C_3(t)) = \left (\frac{1 + t}{2}, \frac{1 – t}{2} \right)
\end{array}
\right \} ~ ~ ~ ~ \text{for} ~ ~ ~ ~ 0 \leq t \leq 1$$
This gives us the lines from $(0,0)$ to $( \frac{1}{2}, \frac{1}{2} )$, from $(0,0)$ to $(1,0)$, and from $( \frac{1}{2}, \frac{1}{2} )$ to $(1,0)$. This is a small triangle (draw it out! :D). It's obvious from looking at this triangle that its area is $\frac{1}{4}$, which should appropriately be changed to $\frac{1}{2}$ by the Jacobian (1st problem: It'd be changed to $-\frac{1}{2}$).
Replacing $x$ and $y$ with $u + v$ and $u-v$ respectively, I get:
$$\iint_D (x + y)~~dxdy = -2\iint 2u~~dudv$$
Since this is a strange triangle, I split it up into two integrals. I take the double integral where $u \in [0, \frac{1}{2}$ and $v \in [0, u]$ and add it to the double integral where $u \in [\frac{1}{2}, 1]$ and $v \in [\frac{1}{2}, 1-u]$. I can also do things like $u \in [0, \frac{1}{2}]$ and $v \in [u, 0]$, etc. (2nd Problem: I have no idea how to pick the bounds of $v$ appropriately; that is, how do I know which one to use when they both give the same line-segment?)
Either way, I don't get the right answer. I'm all messed up.
Help, please. 🙂
Best Answer
Since the region $D:\{(x,y) ~|~ 0 \leq y \leq x, 0 \leq x \leq 1\}$, then the integral is simply $$ \begin{align} \iint_D (x + y)\ dx\ dy&=\int_{x=0}^1\int_{y=0}^x (x+y)\ dy\ dx\\ &=\int_{x=0}^1\left[xy+\frac{1}{2}y^2\right]_{y=0}^x \ dx\\ &=\int_{x=0}^1\left(\frac{3}{2}x^2\right)\ dx\\ &=\left[\frac{1}{2}x^3\right]_{x=0}^1\\ &=\boxed{\color{blue}{\Large\frac{1}{2}}} \end{align} $$ No transformation of variables needed.
UPDATE:
If we want to answer this question by using transformation of variables: $x=u+v$ and $y=u-v$, then the region $D$ in $uv$-coordinate corresponds to the region $$0\le y\le x\;\Rightarrow\;0\le u-v\le u+v\;\Rightarrow\;-v\le v\le u$$ and $$0\le x\le 1\;\Rightarrow\;0\le u+v\le 1.$$ Take a look the picture below.
Thus, $$ \begin{align} \iint_D (x + y)\ dx\ dy&=\iint_D 2u\ |J|\ du\,dv\\ &=4\iint_D u \ dv\,du\\ &=4\left(\int_{u=0}^\frac{1}{2}\int_{v=0}^u u\ dv\,du+\int_{u=\frac{1}{2}}^1\int_{v=0}^{1-u} u\ dv\,du\right)\\ &=4\left(\int_{u=0}^\frac{1}{2} u^2\ du+\int_{u=\frac{1}{2}}^1 u(1-u)\ du\right)\\ &=4\left(\left.\frac{1}{3} u^3\right|_{u=0}^\frac{1}{2}+\left.\frac{1}{2} u^2-\frac{1}{3} u^3\right|_{u=\frac{1}{2}}^1\right)\\ &=4\left(\frac{1}{24}+\frac{3}{8} -\frac{7}{24}\right)\\ &=\boxed{\color{blue}{\Large\frac{1}{2}}} \end{align} $$
$$\\$$
$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$