\( \newcommand{\sech}{\operatorname{sech}} \) \( \newcommand{\inverse}[1]{#1^\leftarrow} \) \( \newcommand{\<}{\langle} \) \( \newcommand{\>}{\rangle} \) \( \newcommand{\vect}{\mathbf} \) \( \newcommand{\veci}{\mathbf{\hat ı}} \) \( \newcommand{\vecj}{\mathbf{\hat ȷ}} \) \( \newcommand{\veck}{\mathbf{\hat k}} \) \( \newcommand{\curl}{\operatorname{curl}\,} \) \( \newcommand{\dv}{\operatorname{div}\,} \) \( \newcommand{\detThree}[9]{ \operatorname{det}\left( \begin{array}{c c c} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \end{array} \right) } \) \( \newcommand{\detTwo}[4]{ \operatorname{det}\left( \begin{array}{c c} #1 & #2 \\ #3 & #4 \end{array} \right) } \)

MA 227 Standard S08


Transformations of Variables

At the end of the course, each student should be able to…

  • S08: TransVar. Compute and apply a transformation of variables.

S08: Transformations of Variables

  • A transformation of variables is a function \(\vect T(u,v)=\<x,y\>\) that converts vectors \(\<u,v\>\) in the \(uv\) plane into vectors \(\<x,y\>\) in the \(xy\) plane.
    • A transformation is called affine if it preserves parallelograms.
    • All affine transformations are of the form \(\vect T(u,v)=\<a_1u+b_1v+c_1,a_2u+b_2v+c_2\>\).
    • To find a transformation from the unit square in the \(uv\) plane to a parallelogram, the values of \(a,b,c\) may be calculated by setting each of \(\vect T(0,0),\vect T(1,0), \vect T(1,1), \vect T(0,1)\) to each of its corners.
    • To find a transformation from the unit triangle in the \(uv\) plane to a parallelogram, the values of \(a,b,c\) may be calculated by setting each of \(\vect T(0,0),\vect T(1,0), \vect T(1,1)\) to each of its corners.
  • Affine transformations scale areas by a factor of \(\detTwo{a_1}{b_1}{a_2}{b_2}\) (where this value is negative when the transformation reflects orientation). This generalizes to the Jacobian \(\frac{\partial\vect T}{\partial\<u,v\>}=\detTwo {\frac{\partial x}{\partial u}}{\frac{\partial x}{\partial v}} {\frac{\partial y}{\partial u}}{\frac{\partial y}{\partial v}}\) for more arbitrary transformations.
    • It follows that if the transformation \(\vect T(u,v)\) transforms the region \(G\) in the \(uv\) plane into the region \(R\) in the \(xy\) plane, then \(\iint_R f(x,y)\,dA=\iint_G f(\vect T(u,v))|\frac{\partial\vect T}{\partial\<u,v\>}|\,dA\).
  • Similarly for 3D transformations \(\vect T(u,v,w)=\<x,y,z\>\) sending \(H\) to \(D\), it may be shown that \(\iiint_D f(x,y,z)\,dV=\iiint_H f(\vect T(u,v,w))|\frac{\partial\vect T}{\partial\<u,v,w\>}|\,dV\) where \(\frac{\partial\vect T}{\partial\<u,v,w\>}=\detThree {\frac{\partial x}{\partial u}}{\frac{\partial x}{\partial v}} {\frac{\partial x}{\partial w}} {\frac{\partial y}{\partial u}}{\frac{\partial y}{\partial v}} {\frac{\partial y}{\partial w}} {\frac{\partial z}{\partial u}}{\frac{\partial z}{\partial v}} {\frac{\partial z}{\partial w}}\).

Textbook References

  • University Calculus: Early Transcendentals (3rd Ed)
    • 14.8

Practice Exercises

Let the unit square have vertices \(\<0,0\>\), \(\<1,0\>\), \(\<1,1\>\), and \(\<0,1\>\). Let the unit triangle have vertices \(\<0,0\>\), \(\<1,0\>\), and \(\<1,1\>\).

  1. Find a transformation from either the unit square or unit triangle in the \(uv\) plane into the given region \(R\) in the \(xy\) plane.
    • \(R\): the parallelogram bounded by \(y=3x+1\), \(y=3x-3\), \(y=x-3\) \(y=x+1\)
    • \(R\): the triangle bounded by \(y=x\), \(y=2x\), \(y=6-x\)
    • \(R\): the square with vertices \(\<2,1\>\), \(\<-2,3\>\), \(\<0,7\>\), \(\<4,5\>\)
    • \(R\): the triangle with vertices \(\<0,-2\>\) \(\<-1,1\>\), \(\<1,3\>\)
  2. Evaluate the double integral with variables \(x,y\) using the given transformation from the \(uv\) plane.
    • \(\iint_R (2y-4x)\,dA\), \(\vect{T}(u,v)=\<u+v,2u-v+3\>\) from the unit square into the parallelogram \(R\) with vertices \(\<0,3\>\), \(\<1,5\>\), \(\<2,4\>\), \(\<1,2\>\).
    • \(\iint_R (x+y)(x-y-2)\,dA\), \(\vect{T}(u,v)=\<4-u-v,v-u+2\>\) from unit triangle into the triangle \(R\) with vertices \(\<4,2\>\), \(\<3,1\>\), \(\<2,2\>\).
    • \(\iint_R (x+y)e^{x^2-y^2}\,dA\), \(\vect{T}(u,v)=\<u+2v,u-2v\>\) from unit square into the rectangle \(R\) bounded by \(y=x\), \(y=x-4\), \(y=-x\), \(y=2-x\).
    • \(\iint_R \cos(e^x)\,dA\), \(\vect{T}(u,v)=\<\ln (u+v+1),v\>\) from unit triangle into the region \(R\) bounded by \(y=0\), \(y=e^x-2\), \(y=\frac{e^x-1}{2}\).

Solutions

  1. Find the transformation (solutions are not unique):
    • \(\vect T(u,v)=\<2u+2v-2,2u+6v-5\>\)
    • \(\vect T(u,v)=\<3u-v,3u+v\>\)
    • \(\vect T(u,v)=\<4u+2v-2,-2u+4v+3\>\)
    • \(\vect T(u,v)=\<u+v-1,-3u+5v+1\>\)
  2. Evaluate the integral:
    • \(9\)
    • \(\frac{3}{2}\)
    • \(\frac{1}{8}(e^8-9)\)
    • \(-\frac{1}{2}\cos(3)+\cos(2)-\frac{1}{2}\cos(1)\)