How would you go about changing the order of integration in a function say ; $$\int_0^8\int_\sqrt[3]{y}^2 f(x,y)~dx~dy$$
$\endgroup$3 Answers
$\begingroup$\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.
$\endgroup$ 3 $\begingroup$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:
$\hspace{3.2cm}$
The curve between the regions is $y=x^3$ or $x=y^{1/3}$.
$\endgroup$ 4 $\begingroup$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:
$\int_{0}^{8}\int_{\sqrt[3]{y}}^{2}f(x,y)dxdy=\int_{0}^{2}\int_{0}^{x^{3}}f(x,y)dydx$.
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