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. :)

  • $\begingroup$ $\large\left(~y \leq x\ \mbox{and}\ x \leq y~\right)\quad\Longrightarrow\quad x = y$. So ???. $\endgroup$ – Felix Marin Apr 6 '14 at 1:31
  • $\begingroup$ Oh, geeze. So sorry about that. I meant $0 \leq x \leq 1$ $\endgroup$ – AmagicalFishy Apr 6 '14 at 2:13

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.


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, $$ \begin{align} \iint_D (x + y)\ dx\ dy&=\iint_D 2u\ |J|\ du\,dv\\ &=4\iint_D u \ dv\,du\\ &=4\left(\int_{u=0}^\frac{1}{2}\int_{v=0}^u u\ dv\,du+\int_{u=\frac{1}{2}}^1\int_{v=0}^{1-u} u\ dv\,du\right)\\ &=4\left(\int_{u=0}^\frac{1}{2} u^2\ du+\int_{u=\frac{1}{2}}^1 u(1-u)\ du\right)\\ &=4\left(\left.\frac{1}{3} u^3\right|_{u=0}^\frac{1}{2}+\left.\frac{1}{2} u^2-\frac{1}{3} u^3\right|_{u=\frac{1}{2}}^1\right)\\ &=4\left(\frac{1}{24}+\frac{3}{8} -\frac{7}{24}\right)\\ &=\boxed{\color{blue}{\Large\frac{1}{2}}} \end{align} $$


$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

  • $\begingroup$ This is pretty close to what I got, but: Isn't the Jacobian Determinant $-2$ and not $2$? And why does the 2nd integral of $v$ go from $0$ to $1-u$ and not from $1-u$ to $0$ (since it starts off at $\frac{1}{2}$, and not zero? $\endgroup$ – AmagicalFishy Apr 6 '14 at 17:42
  • $\begingroup$ @AmagicalFishy If you make a transformation variables, let say from $(x,y)$ to $(u,v)$, then the relation of the function is $$ f(x,y)=g(u,v)\ |J|. $$ You must take the absolute value of the Jacobian. As you can see in the picture above, the region from $0\le u\le\dfrac{1}{2}$ is bounded by lines $v=0$ and $v=u$, then it is obvious the area of the region is $$ \int_{u=0}^\frac{1}{2}\int_{v=0}^u u\ dv\,du. $$ Similarly with the region from $\dfrac{1}{2}\le u\le1$ is bounded by lines $v=0$ and $v=1-u$, then the area of the region is $$ \int_{u=\frac{1}{2}}^1\int_{v=0}^{1-u} u\ dv\,du. $$ $\endgroup$ – Tunk-Fey Apr 6 '14 at 17:53
  • $\begingroup$ @AmagicalFishy Also, if we want to evaluate the area between two curves using integral, we will do it by subtracting the upper curve with the below curve, right? $\endgroup$ – Tunk-Fey Apr 6 '14 at 18:01
  • $\begingroup$ I mean, the Jacobian is: \begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix} The $|J|$ is the determinant of this matrix, which is $-2$, not $2$. I also understand why the bounds of $v$ are $1-u$ and $0$, but why aren't they switched around? Thanks for all your help, by the way. $\endgroup$ – AmagicalFishy Apr 6 '14 at 20:57
  • $\begingroup$ @AmagicalFishy $|J|$ is not only the determinant but also the absolute value. You may check again my statement in your book. Why are they switched around? If you switch, the result of integral will be negative, it is incorrect since you're basically evaluating the volume in the first octant. As what I've already said in my previous comment, in order evaluating the area between two curves in first quadrant, you must substract the top curve with the below curve. Anyway, I'm quite sure my answer is correct. You can confirm this to your friends or after your lecturer discusses this homework. $\endgroup$ – Tunk-Fey Apr 7 '14 at 2:57

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