Calculus III 14.08 Jacobian Variables

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14.08 Jacobian Variables

  • Understand the Jacobian concept.
  • Use a Jacobian to change variables in a double integral.
Carl Gustav Jacobi (1804-1851)
Figure 14.8.1

The Jacobian is named after the German mathematician Carl Gustav Jacobi.[1] Jacobi is known for his work in many mathematical areas, but his interest in integration stemmed from the problem in finding the circumference for an ellipse.

Jacobians

For the single integral

$$\int_{a}^{b} f(x) \:dx $$

changing the variables by substitution and letting \(x=g(u)\), so that \(dx=g^{\prime}(u)du\), will produce

$$\int_{a}^{b} f(x) \:dx =\int_{c}^{d} f(g(u))g^{\prime}(u) \:du $$

where \(a=g(c)\) and \(b=g(d)\). Note the additional \(g^{\prime}(u)\) in the integrand as a byproduct from the variable change. This also occurs with double integrals

$$\int_{R} \int f(x,y) \: dA = \int_{S} \int f(g(u,v),h(u,v)) \color{red}{\underbrace{\color{black}{\begin{vmatrix}\frac{\partial x}{\partial u} \frac{\partial y}{\partial v}-\frac{\partial y}{\partial u} \frac{\partial x}{\partial v}\end{vmatrix}}}_{\color{red}{\text{Jacobian}}}}\: du \: dv $$

where the variable changes

\(x=g(u,v)\) and \(y=h(u,v)\)

introduces a factor called the Jacobian for \(x\) and \(y\) with respect to \(u\) and \(v\) In defining the Jacobian, it is convenient to use the determinant notation described in Definition 14.8.1.

Definition 14.8.1 Jacobian Factor

If \(x=g(u,v)\) and \(y=h(u,v)\), then the Jacobian for \(x\) and \(y\) with respect to \(u\) and \(v\), denoted by \(\partial(x,y)/\partial(u,v)\), is

$$\frac{\partial(x,y)}{\partial(u,v)}= \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}=\frac{\partial x}{\partial u} \frac{\partial y}{\partial v}-\frac{\partial y}{\partial u} \frac{\partial x}{\partial v}.$$

Example 14.8.1 The Jacobian for Cartesian-to-Polar Conversion

Find the Jacobian for the variable change defined by

\(x= r \cos \theta\) and \(y= r \sin \theta.\)

Solution Applying Definition 14.8.1 produces

$$\frac{\partial(x,y)}{\partial(r,\theta)} $$ $$= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{vmatrix}$$
$$=\begin{vmatrix}\cos \theta & -r \sin \theta\\ \sin \theta & r \cos \theta \end{vmatrix}$$
$$= r \cos^{2} \theta + r \sin^{2} \theta= r.$$
Square Half.jpg

\(S\) in the region in the \(r\theta\)-plane that corresponds to \(R\) in the \(xy\)-plane.
Figure 14.8.2

In Example 14.8.1 the variable change from cartesian to polar coordinates for a double integral can be written as

$$\int_{R} \int f(x,y) \: dA $$ $$= \int_{S} \int f(r \cos \theta, r \sin \theta) r \: dr \: d\theta, \: r >0 $$
$$= \int_{S} \int f(r \cos \theta, r \sin \theta)\begin{vmatrix}\partial(x,y) \\ \partial(r,\theta) \end{vmatrix} \: dr \: d\theta$$     Substitute the Jacobian for \(r\)

where \(S\) is the region in the \(r\theta\)-plane that corresponds to the region \(R\) in the \(xy\)-plane, as shown in Figure 14.8.2. This formula extends Theorem 14.3.1 to polar coordinates.

The common case for a variable change is given by a one-to-one transformation \(T\) from a region \(S\) in the \(uv\)-plane to a region \(R\) in the \(xy\)-plane, as denoted by

\(T(u,v) = (x,y) = (g(u,v),h(u,v))\)

where \(g\) and \(h\) have continuous first partial derivatives in the region \(S\). Note the point \((u,v)\) lies in \(S\) and the point \((x,y)\) lies in \(R\). In most cases the intended transformation is one where the region \(S\) is simpler than the region \(R\).

Example 14.8.2 Finding a Variable Change to Simplify a Region

Region \(R\) in the \(xy\)-plane
Figure 14.8.3

Region \(S\) in the \(uv\)-plane
Figure 14.8.4

Let \(R\) be the region bounded by the lines

\(x-2y=0,\)
\(x-2y=-4,\)
\(x+y=4,\)

and

\(x+y=1\)

as shown in Figure 14.8.3. Find a transformation \(T\) from a region \(S\) to \(R\) such that \(S\) is a cartesian region with sides parallel to the \(u\)- or \(v\)-axis.
Solution Let \(u=x+y\) and \(v=x-2y\). Solving this equation system for \(x\) and \(y\) produces \(T(u,v)=(x,y)\), where

$$x=\frac{1}{3} (2u+v) \text{ and } y=\frac{1}{3} (u-v). $$

The four boundaries for \(R\) in the \(xy\)-plane give rise to the following bounds for S in the \(uv\)-plane.

Bounds on the \(xy\)-Plane Bounds on the \(uv\)-Plane
\(x+y=1 \) \(\rightarrow\) \(u=1 \)
\(x+y=4 \) \(\rightarrow\) \(u=4 \)
\(x-2y=0 \) \(\rightarrow\) \(v=0 \)
\(x-2y=-4 \) \(\rightarrow \:\:\:\:\) \(v=-4 \)

The region \(S\) is shown in Figure 14.8.4. Note that the transformation

$$T(u,v)=(x,y)=\left(\frac{1}{3} [2u+v], \frac{1}{3} [u-v] \right) $$

maps the vertices for region \(S\) onto the vertices for region \(R\). For example,

\(T(1,0)\) $$= \left(\frac{1}{3} [2(1)+0], \frac{1}{3} [1-0] \right)$$ $$= \left(\frac{2}{3},\frac{1}{3} \right) $$
\(T(4,0)\) $$=\left(\frac{1}{3} [2(4)+0], \frac{1}{3} [4-0] \right) $$ $$= \left(\frac{8}{3},\frac{4}{3} \right) $$
\(T(4,-4)\) $$=\left(\frac{1}{3} [2(4)-4], \frac{1}{3} [4-(-4)] \right) $$ $$= \left(\frac{4}{3},\frac{8}{3} \right) $$
\(T(1,-4)\) $$= \left(\frac{1}{3} [2(1)-4], \frac{1}{3} [1-(-4)] \right)$$ $$= \left(-\frac{2}{3},\frac{5}{3} \right).$$

Variable Change for Double Integrals

Changing variables can simplify a double integral by making a change to the region \(R\), the integrand \(f(x,y)\), or both.

Theorem 14.8.1 Variable Change for Double Integrals

Figure 14.8.5

Figure 14.8.6

Let \(R\) be a vertically or horizontally simple region in the \(xy\)-plane. Let \(S\) be a vertically or horizontally simple region in the \(uv\)-plane. Let \(T\) from \(S\) to \(R\) be give by \(T(u,v) = (x,y) = (g(u,v),h(u,v))\), where \(g\) and \(h\) have continuous first partial derivatives. Assume that \(T\) is one-to-one except possibly on the boundary for \(S\). If \(f\) is continuous on \(R\), and \(\partial(x,y)/\partial(u,v)\) is nonzero on \(S\), then

$$\int_{R} \int f(x,y) \: dx \: dy = \int_{S} \int f(g(u,v),h(u,v)) \begin{vmatrix} \frac{\partial(x,y)}{\partial(u,v)}\end{vmatrix} \: du \: dv.$$

Proof Consider the case where \(S\) is a rectangular region in the \(uv\)-plane with vertices \((u,v)\), \((u+\Delta u, v)\), \((u+\Delta u, v + \Delta v)\), and \((u,v+\Delta v)\), as shown in Figure 14.8.5. The area for \(S\) is \(S=\Delta u \Delta v, \: \Delta u > 0, \: \Delta v > 0\). The associated \(R\) region is a parallelogram in the \(xy\)-plane with vertices \(M(g(u,v),h(u,v))\), \(N(g(u+\Delta u,v),h(u+\Delta u, v))\), \(P(g(u+\Delta u,v+ \Delta v),h(u+ \Delta u,v + \Delta v))\), and \(Q(g(u,v+\Delta v),h(u,v+ \Delta v))\), as shown in Figure 14.8.6. The area for \(R\), using the vectors \(\overset{\rightharpoonup}{MN}\) and \(\overset{\rightharpoonup}{MQ}\), is

\( \Delta A \approx \| \overset{\rightharpoonup}{MN} \times \overset{\rightharpoonup}{MQ}\|. \)

For small \(\Delta u\) and \(\Delta v\), the partial derivatives for \(g\) and \(h\) with respect to \(u\) can be approximated by

$$g_{u}(u,v) \approx \frac{g(u+\Delta u,v)-g(u,v)}{\Delta u} $$

and

$$h_{u}(u,v) \approx \frac{h(u+\Delta u,v)-h(u,v)}{\Delta u}. $$

Therefore, \(\overset{\rightharpoonup}{MN}\) is approximately

\(\overset{\rightharpoonup}{MN}\) \(=[g(u+\Delta u,v)-g(u,v)]\textbf{i} + [h(u+\Delta u,v)-h(u,v)]\textbf{j}\)
\(\approx [g(u,v)\Delta u]\textbf{i} + [h(u,v)\Delta u]\textbf{j} \)
$$=\frac{\partial x}{\partial u} \Delta u \textbf{i} + \frac{\partial y}{\partial u} \Delta u \textbf{j}.$$

The approximation for \(\overset{\rightharpoonup}{MQ}\) is

$$\overset{\rightharpoonup}{MQ} \approx \frac{\partial x}{\partial v} \Delta v \textbf{i} + \frac{\partial y}{\partial v} \Delta v \textbf{j}.$$

Combining these results to approximate \(\Delta A\) produces

\( \overset{\rightharpoonup}{MN} \times \overset{\rightharpoonup}{MQ} \) $$\approx \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\ \frac{\partial x}{\partial u} \Delta u & \frac{\partial y}{\partial u} \Delta u & 0\\ \frac{\partial x}{\partial v} \Delta v & \frac{\partial y}{\partial v} \Delta v \textbf{j} & 0\end{vmatrix} $$
$$=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix} \Delta u \Delta v \textbf{k}.$$

The Jacobian notation is,

$$ \Delta A \approx \| \overset{\rightharpoonup}{MN} \times \overset{\rightharpoonup}{MQ}\| \approx \begin{vmatrix} \frac{\partial(x,y)}{\partial(u,v)}\end{vmatrix}\: \Delta u \: \Delta v.$$

Because this approximation improves as \(\Delta u\) and \(\Delta v\) approach 0, the limiting case can be written as

$$ dA \approx \| \overset{\rightharpoonup}{MN} \times \overset{\rightharpoonup}{MQ}\| \approx \begin{vmatrix}\frac{\partial(x,y)}{\partial(u,v)}\end{vmatrix}\: du \: dv.$$

Which proves Theorem 14.8.1,

$$\int_{R} \int f(x,y) \: dx \: dy = \int_{S} \int f(g(u,v),h(u,v)) \begin{vmatrix} \frac{\partial(x,y)}{\partial(u,v)}\end{vmatrix} \: du \: dv.$$

Example 14.8.3 Using a Variable Change to Simplify a Region

Figure 14.8.7

Let \(R\) be the region bounded by the lines \(x-2y=0\), \(x-2y=-4\), \(x+y=4\), and \(x+=1\), as shown in Figure 14.8.7. The area for \(R\) is given by

$$\int_{R} \int 3xy \: dA. $$

Simplify the double integral by changing the variables and evaluate it.
Solution The region \(R\) can be simplified by extending the variable change used in Example 14.8.2.

$$x=\frac{1}{3}(2u+v) \text{ and } y=\frac{1}{3}(u-v)$$

The partial derivatives for \(x\) and \(y\) are

$$\frac{\partial x}{\partial u} = \frac{2}{3}, \:\frac{\partial x}{\partial v} = \frac{1}{3}, \: \frac{\partial y}{\partial u} = \frac{1}{3}, $$

and

$$\frac{\partial y}{\partial v} = \frac{1}{3} $$

which implies the Jacobian is

$$\frac{\partial(x,y)}{\partial(u,v)}$$ $$= \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}$$
$$= \begin{vmatrix} \frac{2}{3}& \frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} \end{vmatrix}$$
$$= - \frac{2}{9} - \frac{1}{9} = -\frac{1}{3}.$$

By Theorem 14.8.1,

$$\int_{R} \int 3xy \: dA$$ $$=\int_{S} \int 3 \left[ \frac{1}{3}(2u+v)\frac{1}{3}(u-v)\right]\begin{vmatrix}\frac{\partial(x,y)}{\partial(u,v)}\end{vmatrix} \: dv \: du$$
$$=\int_{1}^{4} \int_{-4}^{0} \frac{1}{9}(2u^{2}-uv-v^{2}) \: dv \: du $$
$$=\frac{1}{9} \int_{1}^{4} \left[ 2u^{2}v- \frac{uv^{2}}{2}- \frac{v^{3}}{3} \right]_{-4}^{0} \: du$$
$$=\frac{1}{9} \int_{1}^{4} \left( 8u^{2} + 8u- \frac{64}{3} \right) \: du $$
$$= \frac{1}{9} \left[ \frac{8u^{3}}{3} + 4u^{2}- \frac{64}{3}u \right]_{1}^{4} = \frac{164}{9}.$$

Example 14.8.4 Using a Variable Change to Simplify an Integrand

Region \(R\) in the \(xy\)-plane.
Figure 14.8.8

Region \(S\) in the \(uv\)-plane.
Figure 14.8.9

Let \(R\) be the region bounded by a square with vertices (0,1), (1,2), (2,1), and (1,0). Evaluate the integral for the area

$$\int_{R} \int (x+y)^{2} \sin^{2} (x-y) \: dA.$$

Solution Not the sides for \(R\) lie on the lines \(x+y=1\), (x-y=1), \(x+y=3\), and \(x-y=-1\), as shown in Figure 14.8.8. Letting \(u=x+y\) and \(v=x-y\) determines the bounds for region \(S\) in the \(uv\)-plane be

\(1 \leqslant u \leqslant 3\) and \(-1 \leqslant v \leqslant 1 \)

as shown in Figure 14.8.9. Solving for \(x\) and \(y\) for \(u\) and \(v\) produces

$$ x = \frac{1}{2}(u+v) \text{ and } y = \frac{1}{2}(u-v).$$

The partial derivatives for \(x\) and \(y\) are

$$\frac{\partial x}{\partial u}= \frac{1}{2}, \: \frac{\partial x}{\partial v}= \frac{1}{2}, \: \frac{\partial y}{\partial u}= \frac{1}{2} \text{, and } \frac{\partial y}{\partial v}= -\frac{1}{2}$$

which implies the Jacobian is

$$\frac{\partial(x,y)}{\partial(u,v)}$$ $$= \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}$$
$$= \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{vmatrix}$$
$$= - \frac{1}{4} - \frac{1}{4} = -\frac{1}{2}.$$

Applying Theorem 14.8.1 produces

$$\int_{R} \int (x+y)^{2} \sin^{2} (x-y) \: dA $$ $$=\int_{-1}^{1} \int_{1}^{3} u^{2} \sin^{2}v \left( \frac{1}{2} \right) \: du \: dv $$
$$= \left. \frac{1}{2} \int_{-1}^{1} (\sin^{2}v) \frac{u^{3}}{3} \right]_{1}^{3} \: dv$$
$$= \frac{13}{3} \int_{-1}^{1} \sin^{2}v \: dv $$
$$= \frac{13}{6} \int_{-1}^{1} (1-\cos 2v) \:dv $$
$$= \frac{13}{6} \left[ v - \frac{1}{2} \sin 2v \right]_{-1}^{1} $$
$$= \frac{13}{6} \left[ 2 - \frac{1}{2} \sin 2 + \frac{1}{2} \sin(-2) \right]$$
$$= \frac{13}{6} (2- \sin 2) \approx 2.363.$$
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Parent Article: Calculus III 14 Multiple Integration