Calculate $\iint_D xy\,\mathrm dx\, \mathrm dy$

Question

Calculate $$\iint_D xy\,\mathrm dx\, \mathrm dy$$ where $D$ is the region bounded by $y = x^2$, $2y = x^2$, $x^2 = y$ and $x^2 = 2y$.

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Can I do it as follows: $$\iint_D xy\,\mathrm dx \,\mathrm dy = \int_{y^2}^{y^2/2}\mathrm dx \int_{x^2}^{x^2/2} xy~ \mathrm dy = \int_{y^2}^{y^2/2}x\left[\frac{y^2}{2}\right]_{x^2}^{x^2/2}\mathrm dx = \int_{y^2}^{y^2/2}\left(\frac{-3x^5}{8}\right)\mathrm dx$$ However, the integral is going to be a function of $y$. I think I'm wrong with this. Or should I divide this region into $4$ regions with four vertical lines passing through the points of intersections?

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Answer 1

At each step of calculating a multivariate integral, you should not have the variables, with respect to which the integral has been simplified, still present. This occurrence is typically due to wrong choice of the integration region. To correctly express the domain,

• write down all the constraints $$ {x^2/2<y<x^2 \\ y^2/2<x<y^2. } $$

• Choose the order of integration. I choose integration first w.r.t. $x$ and then $y$.

• Derive the bound of $x$ variation in terms of $y$ from the constraints of the integration region, $$ { y^2/2<x<y^2 \\ \sqrt y<x<\sqrt {2y} }\implies { \max\{y^2/2,\sqrt y\}<x<\min\{y^2,\sqrt{2y}\} }. $$

• Find the maximal numerical interval of $y$ variation. This process is simple and gives $(1,2)$ here.

• Express the integral in its alternative form as $$ \int_1^2\int_{\max\{y^2/2,\sqrt y\}}^{\min\{y^2,\sqrt{2y}\}}xydxdy . $$

All done!

Answer 2

By making a change of variables: $$ u=\frac{y}{x^2},\qquad v=\frac{x}{y^2} $$ we have that the conditions that define the domain simplify to $$ \frac{1}{2}\leq u\leq 1,\qquad\frac{1}{2}\leq v\leq 1 $$ Let's see that from the transformation rules we have $uv=1/(xy)$ then $$ J^{-1}=\left|\frac{\partial(u,v)}{\partial(x,y)}\right|=\frac{3}{x^2y^2} $$ and $$ J=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{1}{3u^2v^2} $$ so that $$ \int_Dxydxdy=\int_{D'}\frac{1}{uv}\cdot\frac{1}{3u^2v^2}dudv=\frac{1}{3}\int_{1/2}^1\frac{du}{u^3}\int_{1/2}^1\frac{dv}{v^3}=\frac{3}{4} $$