Jacobians and Transformation of Integrals

Introduction to Jacobians

In multivariable calculus, we often encounter situations where we need to change the variables of integration. This transformation can simplify the integrand or the region of integration, making the integral easier to evaluate. The Jacobian is a crucial tool that quantifies how a change of variables affects the area or volume element in multiple dimensions. It is essentially the determinant of the matrix of partial derivatives of the transformation functions.

Consider a transformation from Cartesian coordinates $(x, y)$ to new coordinates $(u, v)$. This can be expressed as a pair of functions: $x = x(u, v)$ $y = y(u, v)$

The Jacobian of this transformation, denoted by $J$ or $\frac{\partial(x, y)}{\partial(u, v)}$, is defined as the determinant of the Jacobian matrix. The Jacobian matrix is formed by the partial derivatives of $x$ and $y$ with respect to $u$ and $v$:

Jacobian Matrix: $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} $$

The Jacobian determinant is then calculated as: $J = \frac{\partial(x, y)}{\partial(u, v)} = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}$

Similarly, if we are transforming from $(u, v)$ to $(x, y)$, the Jacobian is $\frac{\partial(u, v)}{\partial(x, y)} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x}$.

The absolute value of the Jacobian, $|J|$, represents the scaling factor by which an infinitesimal area element $dA = dx \, dy$ in the $(x, y)$ plane is transformed into an infinitesimal area element $du \, dv$ in the $(u, v)$ plane.

Transformation of Integrals

The fundamental theorem for the transformation of double integrals states that if we have a double integral $\iint_R f(x, y) \, dx \, dy$ over a region $R$ in the $xy$-plane, and we introduce a transformation of variables $x = x(u, v)$ and $y = y(u, v)$, then the integral can be rewritten in terms of $u$ and $v$ as:

$\iint_R f(x, y) \, dx \, dy = \iint_{R'} f(x(u, v), y(u, v)) \left| \frac{\partial(x, y)}{\partial(u, v)} \right| \, du \, dv$

Here, $R'$ is the region in the $uv$-plane that corresponds to the region $R$ in the $xy$-plane under the given transformation. The absolute value of the Jacobian, $\left| \frac{\partial(x, y)}{\partial(u, v)} \right|$, is essential because the area element is always positive.

Example 1: Polar Coordinates Transformation

Let's consider the transformation from Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$. The transformation equations are: $x = r \cos \theta$ $y = r \sin \theta$

We need to find the Jacobian $\frac{\partial(x, y)}{\partial(r, \theta)}$. First, let's compute the partial derivatives: $\frac{\partial x}{\partial r} = \cos \theta$ $\frac{\partial x}{\partial \theta} = -r \sin \theta$ $\frac{\partial y}{\partial r} = \sin \theta$ $\frac{\partial y}{\partial \theta} = r \cos \theta$

Now, form the Jacobian matrix: $$ \begin{bmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix} $$

Calculate the determinant: $J = \frac{\partial(x, y)}{\partial(r, \theta)} = (\cos \theta)(r \cos \theta) - (-r \sin \theta)(\sin \theta)$ $J = r \cos^2 \theta + r \sin^2 \theta$ $J = r (\cos^2 \theta + \sin^2 \theta)$ $J = r (1) = r$

The absolute value of the Jacobian is $|J| = |r|$. Since $r$ in polar coordinates typically represents the radial distance, it is non-negative, so $|J| = r$. This means that $dx \, dy = r \, dr \, d\theta$.

Therefore, a double integral in Cartesian coordinates can be transformed into polar coordinates as: $\iint_R f(x, y) \, dx \, dy = \iint_{R'} f(r \cos \theta, r \sin \theta) \, r \, dr \, d\theta$

Example 2: Evaluating an Integral using Polar Coordinates

Evaluate the integral $\iint_D e^{-(x^2+y^2)} \, dx \, dy$ where $D$ is the disk $x^2 + y^2 \leq a^2$.

The region $D$ is a circle centered at the origin with radius $a$. The integrand $e^{-(x^2+y^2)}$ suggests that polar coordinates would simplify the problem.

The transformation is $x = r \cos \theta$, $y = r \sin \theta$, and $dx \, dy = r \, dr \, d\theta$.

The region $D$ in polar coordinates is described by $0 \leq r \leq a$ and $0 \leq \theta \leq 2\pi$.

The integral becomes: $\iint_D e^{-(x^2+y^2)} \, dx \, dy = \int_0^{2\pi} \int_0^a e^{-r^2} \, r \, dr \, d\theta$

We can evaluate this iteratively. First, consider the inner integral with respect to $r$: $\int_0^a e^{-r^2} \, r \, dr$

Let $u = -r^2$. Then $du = -2r \, dr$, so $r \, dr = -\frac{1}{2} du$. When $r=0$, $u=0$. When $r=a$, $u=-a^2$.

The inner integral becomes: $\int_0^{-a^2} e^u \left(-\frac{1}{2} du\right) = -\frac{1}{2} \int_0^{-a^2} e^u \, du = -\frac{1}{2} [e^u]_0^{-a^2}$ $= -\frac{1}{2} (e^{-a^2} - e^0) = -\frac{1}{2} (e^{-a^2} - 1) = \frac{1}{2} (1 - e^{-a^2})$

Now, substitute this back into the outer integral with respect to $\theta$: $\int_0^{2\pi} \frac{1}{2} (1 - e^{-a^2}) \, d\theta$

Since $\frac{1}{2} (1 - e^{-a^2})$ is a constant with respect to $\theta$: $= \frac{1}{2} (1 - e^{-a^2}) \int_0^{2\pi} d\theta = \frac{1}{2} (1 - e^{-a^2}) [\theta]_0^{2\pi}$ $= \frac{1}{2} (1 - e^{-a^2}) (2\pi - 0) = \pi (1 - e^{-a^2})$

So, $\iint_D e^{-(x^2+y^2)} \, dx \, dy = \pi (1 - e^{-a^2})$.

Jacobians in Three Dimensions

The concept of the Jacobian extends to three dimensions. Consider a transformation from Cartesian coordinates $(x, y, z)$ to new coordinates $(u, v, w)$: $x = x(u, v, w)$ $y = y(u, v, w)$ $z = z(u, v, w)$

The Jacobian matrix is a $3 \times 3$ matrix of partial derivatives: $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{bmatrix} $$

The Jacobian determinant, $J = \frac{\partial(x, y, z)}{\partial(u, v, w)}$, is the determinant of this matrix.

$J = \frac{\partial x}{\partial u} \left( \frac{\partial y}{\partial v} \frac{\partial z}{\partial w} - \frac{\partial y}{\partial w} \frac{\partial z}{\partial v} \right) - \frac{\partial x}{\partial v} \left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial w} - \frac{\partial y}{\partial w} \frac{\partial z}{\partial u} \right) + \frac{\partial x}{\partial w} \left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial y}{\partial v} \frac{\partial z}{\partial u} \right)$

For a triple integral $\iiint_V f(x, y, z) \, dx \, dy \, dz$, the transformation rule is: $\iiint_V f(x, y, z) \, dx \, dy \, dz = \iiint_{V'} f(x(u, v, w), y(u, v, w), z(u, v, w)) \left| \frac{\partial(x, y, z)}{\partial(u, v, w)} \right| \, du \, dv \, dw$

Here, $V'$ is the region in the $uvw$-space corresponding to the region $V$ in the $xyz$-space. The absolute value $|J|$ is the scaling factor for the volume element, so $dx \, dy \, dz = |J| \, du \, dv \, dw$.

Example 3: Spherical Coordinates Transformation

Let's consider the transformation from Cartesian coordinates $(x, y, z)$ to spherical coordinates $(\rho, \theta, \phi)$. The standard convention is: $x = \rho \sin \phi \cos \theta$ $y = \rho \sin \phi \sin \theta$ $z = \rho \cos \phi$ where $\rho \geq 0$, $0 \leq \theta \leq 2\pi$, and $0 \leq \phi \leq \pi$.

We need to compute the Jacobian $\frac{\partial(x, y, z)}{\partial(\rho, \theta, \phi)}$. First, calculate the partial derivatives with respect to $\rho$, $\theta$, and $\phi$: $\frac{\partial x}{\partial \rho} = \sin \phi \cos \theta$ $\frac{\partial x}{\partial \theta} = -\rho \sin \phi \sin \theta$ $\frac{\partial x}{\partial \phi} = \rho \cos \phi \cos \theta$ $\frac{\partial y}{\partial \rho} = \sin \phi \sin \theta$ $\frac{\partial y}{\partial \theta} = \rho \sin \phi \cos \theta$ $\frac{\partial y}{\partial \phi} = \rho \cos \phi \sin \theta$ $\frac{\partial z}{\partial \rho} = \cos \phi$ $\frac{\partial z}{\partial \theta} = 0$ $\frac{\partial z}{\partial \phi} = -\rho \sin \phi$

Now, form the Jacobian matrix: $$ \begin{bmatrix} \sin \phi \cos \theta & -\rho \sin \phi \sin \theta & \rho \cos \phi \cos \theta \\ \sin \phi \sin \theta & \rho \sin \phi \cos \theta & \rho \cos \phi \sin \theta \\ \cos \phi & 0 & -\rho \sin \phi \end{bmatrix} $$

Calculate the determinant. Expanding along the third row is often easiest due to the zero: $J = \cos \phi \left| \begin{matrix} -\rho \sin \phi \sin \theta & \rho \cos \phi \cos \theta \\ \rho \sin \phi \cos \theta & \rho \cos \phi \sin \theta \end{matrix} \right| - 0 + (-\rho \sin \phi) \left| \begin{matrix} \sin \phi \cos \theta & -\rho \sin \phi \sin \theta \\ \sin \phi \sin \theta & \rho \sin \phi \cos \theta \end{matrix} \right|$

Let's evaluate the determinants of the $2 \times 2$ matrices: First $2 \times 2$ determinant: $(-\rho \sin \phi \sin \theta)(\rho \cos \phi \sin \theta) - (\rho \cos \phi \cos \theta)(\rho \sin \phi \cos \theta)$ $= -\rho^2 \sin \phi \cos \phi \sin^2 \theta - \rho^2 \sin \phi \cos \phi \cos^2 \theta$ $= -\rho^2 \sin \phi \cos \phi (\sin^2 \theta + \cos^2 \theta) = -\rho^2 \sin \phi \cos \phi$ Second $2 \times 2$ determinant: $(\sin \phi \cos \theta)(\rho \sin \phi \cos \theta) - (-\rho \sin \phi \sin \theta)(\sin \phi \sin \theta)$ $= \rho \sin^2 \phi \cos^2 \theta + \rho \sin^2 \phi \sin^2 \theta$ $= \rho \sin^2 \phi (\cos^2 \theta + \sin^2 \theta) = \rho \sin^2 \phi$

Substitute these back into the expression for $J$: $J = \cos \phi (-\rho^2 \sin \phi \cos \phi) - \rho \sin \phi (\rho \sin^2 \phi)$ $J = -\rho^2 \sin \phi \cos^2 \phi - \rho^2 \sin^3 \phi$ $J = -\rho^2 \sin \phi (\cos^2 \phi + \sin^2 \phi)$ $J = -\rho^2 \sin \phi (1) = -\rho^2 \sin \phi$

The absolute value of the Jacobian is $|J| = |-\rho^2 \sin \phi|$. Since $\rho \geq 0$ and $0 \leq \phi \leq \pi$, $\sin \phi \geq 0$. Thus, $\rho^2 \sin \phi \geq 0$. $|J| = \rho^2 \sin \phi$.

This means that the volume element in spherical coordinates is $dx \, dy \, dz = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi$.

The transformation of a triple integral to spherical coordinates is: $\iiint_V f(x, y, z) \, dx \, dy \, dz = \iiint_{V'} f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \, \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi$

Example 4: Evaluating a Triple Integral using Spherical Coordinates

Evaluate the integral $\iiint_E z \, dV$ where $E$ is the region in the first octant bounded by the sphere $x^2+y^2+z^2 = 1$.

The region $E$ is a portion of a unit sphere in the first octant. The integrand is $z$. This suggests spherical coordinates.

The transformation is $x = \rho \sin \phi \cos \theta$, $y = \rho \sin \phi \sin \theta$, $z = \rho \cos \phi$. The volume element is $dV = dx \, dy \, dz = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi$.

The region $E$ in spherical coordinates is: For the sphere $x^2+y^2+z^2 = 1$, we have $\rho^2 = 1$, so $\rho = 1$. Thus, $0 \leq \rho \leq 1$. For the first octant: $x \geq 0 \implies \rho \sin \phi \cos \theta \geq 0$. Since $\rho \geq 0$ and $\sin \phi \geq 0$, we need $\cos \theta \geq 0$, which means $0 \leq \theta \leq \pi/2$. $y \geq 0 \implies \rho \sin \phi \sin \theta \geq 0$. Since $\rho \geq 0$ and $\sin \phi \geq 0$, we need $\sin \theta \geq 0$, which also means $0 \leq \theta \leq \pi/2$. $z \geq 0 \implies \rho \cos \phi \geq 0$. Since $\rho \geq 0$, we need $\cos \phi \geq 0$, which means $0 \leq \phi \leq \pi/2$.

So, the region $E'$ in spherical coordinates is $0 \leq \rho \leq 1$, $0 \leq \theta \leq \pi/2$, $0 \leq \phi \leq \pi/2$.

The integrand $z$ becomes $\rho \cos \phi$.

The integral becomes: $\iiint_E z \, dV = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 (\rho \cos \phi) (\rho^2 \sin \phi) \, d\rho \, d\phi \, d\theta$ $= \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta$

This is a separable integral. We can evaluate each part: $\int_0^1 \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_0^1 = \frac{1}{4}$ $\int_0^{\pi/2} \cos \phi \sin \phi \, d\phi$. Let $u = \sin \phi$, $du = \cos \phi \, d\phi$. When $\phi=0, u=0$. When $\phi=\pi/2, u=1$. So, $\int_0^1 u \, du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2}$. $\int_0^{\pi/2} d\theta = [\theta]_0^{\pi/2} = \frac{\pi}{2}$

Multiplying these results: $\frac{1}{4} \times \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{16}$

Thus, $\iiint_E z \, dV = \frac{\pi}{16}$.

Generalizations and Properties

The Jacobian is a fundamental concept in change of variables for multiple integrals. It ensures that the differential area or volume elements are correctly scaled during the transformation.

Key properties: 1. If $x=x(u,v)$ and $y=y(u,v)$ are functions, then $\frac{\partial(x, y)}{\partial(u, v)} = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}$. 2. If $u=u(x,y)$ and $v=v(x,y)$ are functions, then $\frac{\partial(u, v)}{\partial(x, y)} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x}$. 3. The Jacobians of inverse transformations are reciprocals: $\frac{\partial(u, v)}{\partial(x, y)} = \frac{1}{\frac{\partial(x, y)}{\partial(u, v)}}$, provided both Jacobians are non-zero. 4. For a composition of transformations, e.g., $x=x(u,v)$ and $u=u(s,t)$, the Jacobian of $x$ with respect to $t$ is the product of the intermediate Jacobians: $\frac{\partial(x, y)}{\partial(s, t)} = \frac{\partial(x, y)}{\partial(u, v)} \frac{\partial(u, v)}{\partial(s, t)}$.

The absolute value of the Jacobian is used because area and volume are inherently positive quantities.

The transformation of integrals is a powerful technique for simplifying complex integrals by changing the coordinate system or variables. The Jacobian is the essential bridge that connects the old and new integral forms. It is crucial for correctly accounting for the change in the differential area or volume element.