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

Finding the image under $T$

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

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

For the side $A\text{:}u=0,0\le v\le 1$ transforms to $x=\text{−}{v}^2,y=0$ so this is the side $A\prime$ that joins $\left(\mathrm{-1},0\right)$ and $\left(0,0\right).$

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

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

All the points in the entire region of the triangle in the $uv\text{-plane}$ are mapped inside the parabolic region in the $xy\text{-plane}\text{.}$

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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 $T\left(u,v\right)=\left(g\left(u,v\right),h\left(u,v\right)\right),$ where $x=g\left(u,v\right)$ and $y=h\left(u,v\right),$ be a one-to-one $C^1$ transformation. We want to see how it transforms a small rectangular region $S,$ $\text{Δ}u$ units by $\text{Δ}v$ units, in the $uv\text{-plane}$ (see the following figure).

Since $x=g\left(u,v\right)$ and $y=h\left(u,v\right),$ we have the position vector $r\left(u,v\right)=g\left(u,v\right)i+h\left(u,v\right)j$ of the image of the point $\left(u,v\right).$ Suppose that $\left(u_0,v_0\right)$ is the coordinate of the point at the lower left corner that mapped to $\left(x_0,y_0\right)=T\left(u_0,v_0\right).$ The line $v=v_0$ maps to the image curve with vector function $\text{r}\left(u,v_0\right),$ and the tangent vector at $\left(x_0,y_0\right)$ to the image curve is

Similarly, the line $u=u_0$ maps to the image curve with vector function $r\left(u_0,v\right),$ and the tangent vector at $\left(x_0,y_0\right)$ to the image curve is

Now, note that

Similarly,

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

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