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How would you go about changing the order of integration in a function say ; $$\int_0^8\int_\sqrt[3]{y}^2 f(x,y)~dx~dy$$

Michael Hardy
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user90426
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3 Answers3

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\begin{eqnarray*} \int_{0}^{8}\int_{\sqrt[3]{\vphantom{\large a}y\,}}^{2}{\rm f}\left(x, y\right)\,{\rm d}x\,{\rm d}y & = & \int_{0}^{8}\left\lbrack\int_{0}^{2}\Theta\left(x - \sqrt[3]{\vphantom{\large a}y\,} \right) {\rm f}\left(x, y\right)\,{\rm d}x\right\rbrack{\rm d}y \\ & = & \int_{0}^{2}\left\lbrack\int_{0}^{8}\Theta\left(x^{3} - y\right) {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x \\ & = & \int_{0}^{2}\left\lbrack\int_{0}^{x^{3}} {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x \end{eqnarray*}

where Θ is the Heaviside step function.

GKFX
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Felix Marin
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The region of integration is the part of $[0,2]\times[0,8]$ where $x^3\ge y$: $$ \int_0^8\int_{y^{1/3}}^2f(x,y)\,\mathrm{d}x\,\mathrm{d}y =\int_0^2\int_0^{x^3}f(x,y)\,\mathrm{d}y\,\mathrm{d}x $$ The shaded region is where the integration takes place:

$\hspace{3.2cm}$enter image description here

The curve between the regions is $y=x^3$ or $x=y^{1/3}$.

robjohn
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    right but how do you know that? – user90426 Aug 15 '13 at 02:28
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    $y$ goes from $0$ to $8$. $x$ goes from $y^{1/3}$ to $2$. The minimum for $x$ is $0$ (when $y=0$) and the maximum of $x$ is $2$. Thus, all of the $(x,y)$ are in $[0,2]\times[0,8]$, but we also have the further constraint that $x\ge y^{1/3}$. You just read it off of the limits of integration. – robjohn Aug 15 '13 at 02:35
  • By the way, is this the reason why $$\int_0^8\int_{y^{1/3}}^2f(x,y),\mathrm{d}x,\mathrm{d}y \neq \int_0^8\int_0^{y^{1/3} }f(x,y),\mathrm{d}x,\mathrm{d}y$$

    Since in the second case, $x^3 \leq y$?

     – Cheng Mar 18 '21 at 04:46
  • Well, the integral on the left is for the shaded region, whereas the integral on the right is for the white region. Whether they are equal depends on $f$. – robjohn Mar 18 '21 at 07:37
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I find it helps to draw the region you are integrating over when trying to change the order of integration. For this case switching the integrals will give:

$\int_{0}^{8}\int_{\sqrt[3]{y}}^{2}f(x,y)dxdy=\int_{0}^{2}\int_{0}^{x^{3}}f(x,y)dydx$.

user71352
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    ok but how did you get that. – user90426 Aug 15 '13 at 02:24
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    The portion you are integrating over is between the graph of the the function $y=x^{3}$ the vertical line $x=2$ and the horizontal line $y=0$. When integrating over $x$ first we look at slits parallel to the $x$-axis of this region which are $\sqrt[3]{y}\le x\le2$ then we move $y$ from $0$ to $8$. If instead we look at the slits parallel to the $y$-axis then we see the region $0\le y\le x^{3}$ and we move this slit from $x=0$ to $x=2$. – user71352 Aug 15 '13 at 02:30