3.13 Change of Variables

Recall: A change of variables can be useful in double integrals, just like substitution in one - dimensional calculus. \[\iint \limits _R f(x,y)\,dA = \iint \limits _S f(r\cos \theta , r \sin \theta )\, r\,dr\,d\theta \] where \(S\) is the region in the \(r\,\theta - \) plane that corresponds to the region \(R\) in the \(xy-\) plane.

More generally, we consider a change of variables that is given by a “ Transformation \(T\) ” from the \(uv -\) plane to the \(xy-\) plane \[T(uv) = (x,y)\] where \(x = g(u,v) \hspace {0.2cm} y = h(u,v)\) or \(x = x (u,v)\hspace {0.2cm} y = y(u,v)\)

We assume that \(T\) is a \(C^1\) transformation.

\(\implies \) \(h\) and \(g\) have first partial derivatives.

\(T\) transforms \(S\) into the region \(R\) called image of \(S\), consisting of all images of all points in \(S\). It \(T^{-1}\) exists \[u = G(x,y)\hspace {0.2cm}, \hspace {0.3cm} v = H(x,y)\]

The vector \(\displaystyle {r(u,v) = g(u,v) \textbf {i} + h(u,v)\textbf {j}}\)

\begin {align*} r_u & = g_u(u_0,v_o)\textbf {i} + h_u ( u_0,v_0)\textbf {j}\\ & = \frac {\partial x}{\partial u}\textbf {i} + \frac {\partial y}{\partial u}\textbf {j}\\ \end {align*}

\begin {align*} r_v & = g_v(u_0,v_o)\textbf {i} + h_v ( u_0,v_0)\textbf {j}\\ & = \frac {\partial x}{\partial v}\textbf {i} + \frac {\partial y}{\partial v}\textbf {j} \end {align*}

\(a = r(u_0 + \Delta u , v_0 ) - r(u_0, v_0)\)

\(b = r(u_0, v_0 + \Delta v) - r(u_0,v_0)\)

\[\text {But}\hspace {0.4cm} r_u = \lim _{\Delta u \rightarrow 0} \frac {r(u_0 + \Delta u, v_0) - r(u_0,v_0)}{\Delta u}\]

\[\therefore \hspace {0.3cm} r_u \Delta u \approx r(u_0 + \Delta u, v_0) - r(u_0,v_0)\]

\[||||_y \hspace {0.3cm} r_v \Delta v \approx r(u_0 , v_0 + \Delta v) - r(u_0,v_0)\]

We can approximate the area of \(R\) by the area of the parallelogram \[ \big | \big (r_u\Delta u\big ) \times \big ( r_v \Delta \big ) = \big | r_u \times r_v \big |\,\Delta v \,\Delta u\]

\begin {align*} r_u \times r_v & = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\\\ \dfrac {\partial x}{\partial u} & \dfrac {\partial y}{\partial u} & 0\\\\ \dfrac {\partial x}{\partial v} & \dfrac {\partial y}{\partial v} & 0\\ \end {vmatrix}\\\\ & = k \begin {vmatrix} \dfrac {\partial x}{\partial u} & \dfrac {\partial x}{\partial v}\\\\ \dfrac {\partial y}{\partial u} & \dfrac {\partial y}{\partial v}\\ \end {vmatrix}\\\\\\\\\\ \end {align*}

The Jacobian is the two-variable answer to a question already familiar from one variable. In the substitution \(\int f(x)\,dx = \int f\big (g(u)\big )g'(u)\,du\) the factor \(g'(u)\) appears because a small interval \(\Delta u\) is stretched into an interval of length roughly \(\left |g'(u)\right |\Delta u\). In the plane a small rectangle \(\Delta u\,\Delta v\) is carried by \(T\) into a small parallelogram, and the factor by which its area is stretched is the absolute value of the Jacobian: \[\Delta A \approx \left |\frac {\partial ( x,y)}{\partial (u,v)}\right | \Delta u\, \Delta v ,\] the Jacobian being evaluated at \((u_0,v_0)\). It is an area magnification factor, and nothing more mysterious than that. Its sign records whether \(T\) preserves or reverses orientation, which is why the absolute value is taken.

Summing over the rectangles of a partition, \begin {align*} \iint \limits _R f(x,y)\, dA & \approx \sum ^m_{i=1}\sum ^n_{j = 1} f(x_i , y_j)\, \Delta A\\\\ & \approx \sum ^m_{i=1}\sum ^n_{j = 1}f\big (g(u,v), h(u,v)\big ) \left |\frac {\partial ( x,y)}{\partial (u,v)}\right |\, \Delta u \,\Delta v , \end {align*}

and the right-hand side is a Riemann sum for an integral over \(S\). Passing to the limit gives the result the whole construction was for.

The condition \(J\neq 0\) is not decoration: where the Jacobian vanishes the transformation collapses area, and the formula has nothing to say.

Example 3.13.4.

Use the change of variables \(x = u^2 - v^2 \hspace {0.2cm}, \hspace {0.2cm} y = 2uv\hspace {0.2cm}\) to evaluate the integral \[\iint \limits _R y\,dA\] , where \(R\) is the region bounded by the \(x-\) axis and the parabolas \(y^2 = 4 - 4x\) and \(y^2 = 4 + 4x\).

\(T(R) = S\) where \(S\) is the square \([0,1]\times [0,1]\hspace {0.3cm} S\) is a much easier region than \(R\).

\begin {align*} \iint \limits _R y\,dA & = \iint \limits _S 2uv \,\big |4u^2 + 4v^2\big |\, du\, dv\\ & =\int ^1_0\int ^1_0 \big (8u^3v + 8uv^3\big )\,du\,dv\\ & = 2 \end {align*}

\[ \text {Check}\hspace {0.5cm} \iint \limits _R y\, dA = \int ^0_{-1}\int ^{\sqrt {4 + 4x}}_0 \,y \,dy \,dx\]

Example 3.13.5.

Evaluate the integral \(\displaystyle {\iint \limits _ R e^{(x+y)/(x-y)}dA}\hspace {0.2cm}\) where \(R\) is the trapezoidal region with vertices \((1,0)\hspace {0.1cm}, \hspace {0.1cm} (2,0) \hspace {0.1cm},\hspace {0.1cm} (0,-2)\hspace {0.1cm}, \hspace {0.1cm} (0,-1)\).

Let \(u = x + y \) and \(v = x - y\)

\(T^{-1}\) from the \(xy\) plane to the \(uv\) plane \(T:\hspace {0.3cm} x = \dfrac {1}{2}(u+ v) \hspace {0.4cm} y = \frac {1}{2}(u- v)\)

\[S = \big \{ (u,v)\hspace {0.1cm} \big | \hspace {0.1cm} 1\leq v \leq 2 \hspace {0.1cm}, \hspace {0.1cm} -v \leq u \leq v\big \}\]

\begin {align*} \iint \limits _ R e^{(x+y)/(x-y)}\,dA & = \iint \limits _S e^{u/v}\,\big |J(u,v)\big | \,du \,dv\\ & = \int ^2_1 \int ^v_{-v} e^{u/v} \cdot \frac {1}{2}\,du\, dv\\\\ & = \frac {1}{2}\int ^2_1 \Big ( e - e^{-1}\Big ) v \hspace {0.1cm}dv\\\\ & = \frac {3}{4}\Big ( e - e^{-1}\Big )\\ \end {align*}