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Separable Integrals

April 12th, 2009 |

Author: Merrin

Separable integrals are multiple integrals that can be calculated as the product of single integrals. The calculation will come out the same whether from a product of single integrals or an iterated integral. The requirement on the function being integrated is that it decomposes into a product of functions of the integration variables. Variable limits of integration also wreck separability because variables from one integration feed into the next. Recall that variable limits look like

$\displaystyle \int _{x_{1} }^{x_{2} }\int _{y_{1} (x)}^{y_{2} (x)}f(x,y)dydx$

The integration for $latex f(x,y)=g(x)h(y)$ can be carried out as follows over the rectangle $latex a\le x\le b\, , c\le y\le d$.

$\displaystyle {\int _{a}^{b}\int _{c}^{d}f(x,y)dydx }={\int _{a}^{b}\int _{c}^{d}g(x)h(y)dydx =\left(\int_a^bg(x)dx\right) \left(\int_c^df(y)dy\right)}$

Integrals that are not separable in one coordinate system may become separable in another. The symmetry of the coordinate system dictates whether an integral is separable or not. For example in Cartesian coordinates, the area of a square can be expressed as a separable integral.

$\displaystyle A_{square} =\int _{0}^{L}\int _{0}^{L}dydx=\left(\int _{0}^{L}ds \right)^{2} =L^{2}$

The volume of a cube represented as an integral in Cartesian coordinates and is separable.

$\displaystyle V_{cube} =\int _{0}^{L}\int _{0}^{L}\int _{0}^{L}dzdydx$

$=\left(\int _{0}^{L}dx \right)\left(\int _{0}^{L}dy \right)\left(\int _{0}^{L}dz \right)=L^{3}$

The area of a circle in Cartesian coordinates is not separable

$\displaystyle A_{circle} =\int _{-R}^{R}\int _{-\sqrt{R^{2} -x^{2} } }^{\sqrt{R^{2} -x^{2} } }dydx$

Variable limits of integration kill the separability. Later we will show however in polar coordinates the area of a circle can be expressed as a separable integral.

$\displaystyle A_{circle} =\int _{0}^{2\pi }\int _{0}^{R}rdrd\theta =\int _{0}^{2\pi }d\theta \int _{0}^{R}rdr= (2\pi )(\cfrac{1}{2} R^{2} )=\pi R^{2}$

The volume of a sphere in Cartesian coordinates is also not a separable integral because of variable limits of integration.

$\displaystyle V_{sphere} =\int _{x=-R}^{R}\int _{y=-\sqrt{R^{2} -x^{2} } }^{\sqrt{R^{2} -x^{2} } }\int _{z=-\sqrt{R^{2} -x^{2} -y^{2} } }^{\sqrt{R^{2} -x^{2} -y^{2} } }dzdydx$

But in spherical coordinates, we will show how to write the volume of a sphere as

$\displaystyle V_{sphere}=\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }\int _{r=0}^{R}r^{2} \sin \theta d\theta drd\phi$

$\displaystyle V_{sphere}=\int _0^{2\pi }d\phi \int _{0}^{\pi }\sin \theta d\theta \int _0^Rr^{2} dr =\cfrac{4}{3}\pi R^3$

Example 1. Evaluate the integral of $latex y\ln e^{xy} $ over the rectangle defined by

$R:0\le x\le 1,0\le y\le 2$

Solution 1. First set up the proper limits of integration with the function in the integrand.

$\displaystyle \int _{0}^{1}\int _{0}^{2}y\ln e^{xy} dydx$

We simplify the integrand and realize the integral is separable. We can do these simple single definite integrals

in our head.

$\displaystyle \int _{0}^{1}\int _{0}^{2}xy^{2} dydx =\int _{0}^{1}xdx \int _{0}^{2}y^{2} dy =\cfrac{1}{2} \cfrac{8}{3} =\cfrac{4}{3}$

Example 2. Reduce the following multiple integral to a calculation of a single integral.

$\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2} -y^{2} -z^{2} } dzdydx$

Solution 2. This integral meets the requirements for being separable. The function is a product of functions of each independent variable. We will show later how to evaluate the single integral by using a change of coordinate systems.

$\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2} -y^{2} -z^{2} } dzdydx=\left(\int _{-\infty }^{\infty }e^{-x^{2} } dx \right) ^{3} =I^{3} =\pi ^{3/2}$

When dealing with higher order multiple integrals, it may turn out that some of the variables are separable while others are connected. In this case, the integral is still partially separable. For example let

$f(x,y,z)=g(x)h(y,z)$

$\displaystyle {\int _{x_{1} }^{x\_ _{2} }\int _{y_{1} }^{y_{2} }\int _{z_{1} }^{z_{2} }f(x,y,z)dzdydx} = {\int _{x_{1} }^{x_{2} }\int _{y_{1} }^{y_{2} }\int _{z_{1} }^{z_{2} }g(x)h(y,z)dzdydx } \\ =\int _{x_{1} }^{x_{2} }g(x)dx\int _{y_{1} }^{y_{2} }\int _{z_{1} }^{z_{2} }h(y,z)dzdy$

The x integral is separable but the y and z integrals are linked. Certain integrals can be presented in a form where they are not separable but can be made so by using an identity or simplification.

Example 3. Evaluate the following multiple integral, by converting it into a separable form

$\displaystyle \int _{0}^{\pi /2}\int _{0}^{\pi /2}\cos (x+2y)dxdy$

Solution 3. To calculate with separable integrals we recall that

$\cos (x+2y)=\cos (x)\cos (2y)-\sin (x)\sin (2y)$

$\displaystyle {I} = {\int _{0}^{\pi /2}\cos (x)\int _{0}^{\pi /2}\cos (2y) -\int _{0}^{\pi /2}\sin (x)\int _{0}^{\pi /2}\sin (2y) }$

$\displaystyle I = \left( \sin (x)|_{0}^{\pi/2} \right) \left( \cfrac{\sin 2y}{2}|_{0}^{\pi/2} \right) -\left(-\cos x |_0^{\pi/2} \right) \left( -\cfrac{\cos (2x)} {2} |_0^{\pi/2} \right)$