Part 4: Change of Variable for Numerical Purposes

Many algorithms for approximating double integals numerically require the use of special types of regions and sophisticated methods for partitioning those regions. In such instances, the change of variables formula is often used as a tool for transforming a double integral into a form that is more amenable to numerical approximation, such as converting to a rectangle over a rectangular region since a rectangle can be easily partitioned.

Most importantly, however, a change of variable might lead to the reduction of a double integral to a single integral, in which case only a single integral need be approximated numerically. In general, numerical methods for single integrals are preferable to numerical methods for multiple integrals.      

EXAMPLE 5    Transform the following using x = vcosh(u) , y = vsinh( u) :

ó õ Ö3 0  ó õ Ö y2+9 2y      sin(x2-y2) x2-y2 dydx

Evaluate one of the integrals, and then approximate the remaining integral numerically.       

Solution: To do so, we notice that as a double integral we have

ó õ Ö3 0  ó õ Ö y2+9 2y   sin(x2-y2) x2-y2 dxdy = ó õ ó õ R   sin(x2-y2) x2-y2 dA

where R is the region bounded by y = 0, x = 2y, and x² = y²+9. The pullback of R to S is accomplished by letting x = vcosh(u) and y = vsinh( u) :

y = 0    Þ     vsinh( u) = 0   Þ     v = 0  or  sinh( u) = 0    Þv = 0  or  u = 0 x = 2y    Þ     vcosh( u) = 2vsinh(u)     Þ     tanh( u) = 0.5   Þ     u = tanh-1( 0.5) x2 = y2+9    Þ     v2cosh2( u)-v2sinh2( u) = 9    Þ     v2 = 9

Thus, the pullback of R is u = 0, u = tanh^-1( 2) , v = 0, v = 3.

            We next compute the Jacobian determinant:

¶( x,y) ¶( u,v) =  ¶x ¶u  ¶y ¶v -  ¶x ¶v  ¶y ¶u = vsinh( u) sinh( u) -cosh( u)vcosh( u) = -v( cosh2( u) -sinh2( u) ) = -v

Thus, dA = | -v| dudv = vdudv since v is inside of S. Since x²-y² = v², the double integral becomes

ó õ ó õ R   sin( x2-y2) x2-y2 dA = ó õ 3 0  ó õ tanh-1( 0.5) 0   sin(v2) v2 vdudv = ó õ 3 0      sin( v2) v u ê ê tanh-1( 0.5) 0    dv = tanh-1( 0.5) ó õ 3 0   sin( v2) v dv

Finally, the remaining integral can be estimated numerically so that

ó õ Ö3 0  ó õ Ö y2+9 2y   sin(x2-y2) x2-y2 dydx = tanh-1( 0.5) ó õ 3 0   sin( v2) v dv » 0.4573