Evaluating the double integral $\int\int_{R}xy\:dA$ using change of variable $ux=y,vx=1$

Question

Consider the region $R$ bounded by the lines $y=x,y=2x,x+y=2$. Find $$\int\int_{R}xydxdy$$

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I actually used Change of variable using the following transformation: $u=\frac{y}{x}$ and $v=\frac{1}{x}$. We have $1\leq u \le 2$ and $x+y=2$ gives $1+u=2v$

So the region in the $u-v$ plane is $$R_{uv}=\left\{(u,v):1\leq u \leq 2, \frac{1+u}{2}\leq v\leq \frac{3}{2}\right\}$$ I know that my limits are wrong because I am not getting the correct answer as $\frac{13}{81}$. Can I know whats wrong in my transformation?

Answer 1

As other answers have said, you do not have to use change of variable. If you are using change of variable, I would have gone with, $u = \cfrac{y}{x}, v = y$, and that leads to

$|J^{-1}| = \cfrac{y}{x^2}$. So the integrand becomes,

$x y \cdot |J| = x^3 = \cfrac{v^3}{u^3}$

$0 \leq v \leq \frac{2u}{u+1}, 1 \leq u \leq 2$

For change of variable that you are using,

$u = \frac{y}{x}, v = \frac{1}{x}$,

$|J^{-1}| = \cfrac{1}{x^3}$. So the integrand becomes,

$|J| \cdot xy = x^4 y = \cfrac{u}{v^5}$

Please note when $x = 0, v = \infty$ and that is the upper bound of $v$. Lower bound of $v$ comes from $x + y = 2$

$\frac{1+u}{2} \leq v \leq \infty, 1 \leq u \leq 2$

$ \displaystyle \int_1^2 \int_{(1+u)/2}^{\infty} \frac{u}{v^5} ~ dv ~ du = \cfrac{13}{81}$

Answer 2

Without using change of variables: $$\underbrace{\iint_Rxydxdy}_{f\vert_R\text{ is measurable and non-negative}}=\int_0^{\frac{2}{3}}xdx\int_x^{2x}ydy+\int_{\frac{2}{3}}^1xdx\int_x^{2-x}ydy=\\\int_0^{\frac 2 3}x\cdot\dfrac{y^2}{2}\bigg\vert_{x}^{2x}dx+\int_{\frac 23}^1x\cdot \dfrac{y^2}{2}\bigg\vert_{x}^{2-x}=\\=\int_0^{\frac 23}\dfrac{3x^3}{2}dx+\int_{\frac 23}^1 2x-2x^2dx.$$

Answer 3

You did not find the correct limits after the transformation, and you don't need to make the transformation. We have

$$\int\int_{R}xydxdy=\int_{0}^{\frac{2}{3}}\int_{x}^{2x}xy \ dydx+\int_{\frac{2}{3}}^{1}\int_{x}^{2-x} xy \ dydx$$

and after the transformation $y=\frac{u}{v}$ and $x=\frac{1}{v}$ we obtain

$$\int_{\frac{3}{2}}^{\infty}\int_{1}^{2}\frac{u}{v^2}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|dudv+\int_{1}^{\frac{3}{2}}\int_{1}^{2v-1}\frac{u}{v^2}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|dudv$$ $$=\int_{\frac{3}{2}}^{\infty}\int_{1}^{2}\frac{u}{v^5}dudv+\int_{1}^{\frac{3}{2}}\int_{1}^{2v-1}\frac{u}{v^5}dudv=\frac{2}{27}+\frac{7}{81}=\frac{13}{81}$$