5.7 Change of variables in multiple integrals (Page 2/13)

Calculus volume 3 Page 2 / 13

Finding the image under $T$

Let the transformation $T$ be defined by $T (u, v) = (x, y)$ where $x = u^{2} - v^{2}$ and $y = u v .$ Find the image of the triangle in the $u v -plane$ with vertices $(0, 0), (0, 1),$ and $(1, 1) .$

The triangle and its image are shown in [link] . To understand how the sides of the triangle transform, call the side that joins $(0, 0)$ and $(0, 1)$ side $A,$ the side that joins $(0, 0)$ and $(1, 1)$ side $B,$ and the side that joins $(1, 1)$ and $(0, 1)$ side $C .$

On the left-hand side of this figure, there is a triangular region given in the Cartesian uv-plane with boundaries A, B, and C represented by the v axis, the line u = v, and the line v = 1, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u squared minus v squared and y = u v. On the right-hand side of this figure there is a complex region given in the Cartesian x y-plane with boundaries A’, B’, and C’ given by the x axis, y axis, and a line curving from (negative 1, 0) through (0, 1), namely x = y squared minus 1, respectively. — A triangular region in the $u v -plane$ is transformed into an image in the $x y -plane .$

For the side $A : u = 0, 0 \leq v \leq 1$ transforms to $x = − v^{2}, y = 0$ so this is the side $A'$ that joins $(−1, 0)$ and $(0, 0) .$

For the side $B : u = v, 0 \leq u \leq 1$ transforms to $x = 0, y = u^{2}$ so this is the side $B^{'}$ that joins $(0, 0)$ and $(0, 1) .$

For the side $C : 0 \leq u \leq 1, v = 1$ transforms to $x = u^{2} - 1, y = u$ (hence $x = y^{2} - 1)$ so this is the side $C^{'}$ that makes the upper half of the parabolic arc joining $(−1, 0)$ and $(0, 1) .$

All the points in the entire region of the triangle in the $u v -plane$ are mapped inside the parabolic region in the $x y -plane .$

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Let a transformation $T$ be defined as $T (u, v) = (x, y)$ where $x = u + v, y = 3 v .$ Find the image of the rectangle $G = {(u, v) : 0 \leq u \leq 1, 0 \leq v \leq 2}$ from the $u v -plane$ after the transformation into a region $R$ in the $x y -plane .$ Show that $T$ is a one-to-one transformation and find $T^{−1} (x, y) .$

$T^{−1} (x, y) = (u, v)$ where $u = \frac{3 x - y}{3}$ and $v = \frac{y}{3}$

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Jacobians

Recall that we mentioned near the beginning of this section that each of the component functions must have continuous first partial derivatives, which means that $g_{u}, g_{v}, h_{u},$ and $h_{v}$ exist and are also continuous. A transformation that has this property is called a $C^{1}$ transformation (here $C$ denotes continuous). Let $T (u, v) = (g (u, v), h (u, v)),$ where $x = g (u, v)$ and $y = h (u, v),$ be a one-to-one $C^{1}$ transformation. We want to see how it transforms a small rectangular region $S,$ $Δ u$ units by $Δ v$ units, in the $u v -plane$ (see the following figure).

On the left-hand side of this figure, there is a region S with lower right corner point (u sub 0, v sub 0), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with point (x sub 0, y sub 0) given in the Cartesian x y-plane with sides r(u, v sub 0) along the bottom and r(u sub 0, v) along the left. — A small rectangle $S$ in the $u v -plane$ is transformed into a region $R$ in the $x y -plane .$

Since $x = g (u, v)$ and $y = h (u, v),$ we have the position vector $r (u, v) = g (u, v) i + h (u, v) j$ of the image of the point $(u, v) .$ Suppose that $(u_{0}, v_{0})$ is the coordinate of the point at the lower left corner that mapped to $(x_{0}, y_{0}) = T (u_{0}, v_{0}) .$ The line $v = v_{0}$ maps to the image curve with vector function $r (u, v_{0}),$ and the tangent vector at $(x_{0}, y_{0})$ to the image curve is

r_{u} = g_{u} (u_{0}, v_{0}) i + h_{u} (u_{0}, v_{0}) j = \frac{\partial x}{\partial u} i + \frac{\partial y}{\partial u} j .

Similarly, the line $u = u_{0}$ maps to the image curve with vector function $r (u_{0}, v),$ and the tangent vector at $(x_{0}, y_{0})$ to the image curve is

r_{v} = g_{v} (u_{0}, v_{0}) i + h_{v} (u_{0}, v_{0}) j = \frac{\partial x}{\partial v} i + \frac{\partial y}{\partial v} j .

Now, note that

r_{u} = lim_{Δ u \to 0} \frac{r (u_{0} + Δ u, v_{0}) - r (u_{0}, v_{0})}{Δ u} so r (u_{0} + Δ u, v_{0}) - r (u_{0}, v_{0}) \approx Δ u r_{u} .

Similarly,

r_{v} = lim_{Δ v \to 0} \frac{r (u_{0}, v_{0} + Δ v) - r (u_{0}, v_{0})}{Δ v} so r (u_{0}, v_{0} + Δ v) - r (u_{0}, v_{0}) \approx Δ v r_{v} .

This allows us to estimate the area $Δ A$ of the image $R$ by finding the area of the parallelogram formed by the sides $Δ v r_{v}$ and $Δ u r_{u} .$ By using the cross product of these two vectors by adding the k th component as $0,$ the area $Δ A$ of the image $R$ (refer to The Cross Product ) is approximately $| Δ u r_{u} \times Δ v r_{v} | = | r_{u} \times r_{v} | Δ u Δ v .$ In determinant form, the cross product is

r_{u} \times r_{v} = | \begin{matrix} i & j & k \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & 0 \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & 0 \end{matrix} | = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{array} | k = (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) k .

Since $| k | = 1,$ we have $Δ A \approx | r_{u} \times r_{v} | Δ u Δ v = (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) Δ u Δ v .$

Definition

The Jacobian of the $C^{1}$ transformation $T (u, v) = (g (u, v), h (u, v))$ is denoted by $J (u, v)$ and is defined by the $2 \times 2$ determinant

J (u, v) = | \frac{\partial (x, y)}{\partial (u, v)} | = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{array} | = (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) .

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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