Part 13 - The Jacobian | rezarezvan.com

Introduction

In this part we’ll cover how we change our variables into others, in the most general case.

General case of variable change

\iint_D f(x, y)\ dA

We want to rewrite this in terms of $u$ and $v$ . Let’s say that:

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

We can say that we have a transformation, $T$ , from the $uv$ -plane to the $xy$ -plane, given by our functions $g$ and $h$ .

We need to make some assumptions of $T$ to make our lives easier.

$T$ is a $C^1$ -transformation, meaning that $g$ and $h$ have continuous partial derivatives.
$T$ is an injective transformation (meaning it is 1-to-1). This means that we can express $u$ and $v$ in terms of $x$ and $y$ .

With this, we can define the Jacobian matrix.

Definition 1 (Jacobian)

The jacobian matrix of a transformation, $T$ , is:

\dfrac{\partial(x, y)}{\partial(u, v)} = det \begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \newline \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{bmatrix}

Theorem 1 (Change of variables)

Suppose $T$ is a transformation, from the $uv$ -plane to the $xy$ -plane. Assuming that the jacobian for $T$ , is non-zero, then:

\iint_D f(x, y)\ dA = \iint_S f(x(u, v), y(u, v)) \vert \tfrac{\partial(x, y)}{\partial(u, v)} \vert \ dA

Example 1

Find

\iint_D e^{\frac{x + y}{x - y}}\ dA

Where $D$ is the trapezoidal region, with vertices, $(1, 0), (2, 0), (0, -2)$ and $(0, -1)$ .

Let $u = x + y$ and $v = x - y$ .

This means that:

x = \frac{u + v}{2} \newline y = \frac{u - v}{2}

Let’s compute the jacobian:

\dfrac{\partial(x, y)}{\partial(u, v)} = det \begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \newline \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{bmatrix}

\dfrac{\partial(x, y)}{\partial(u, v)} = det \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \newline \frac{1}{2} & -\frac{1}{2} \end{bmatrix}

\dfrac{\partial(x, y)}{\partial(u, v)} = -\frac{1}{2}

Since our transformation is linear, we only need to recompute our vertices. Using our formulas we find that

(1, 0) \rarr (1, 1) \newline (2, 0) \rarr (2, 2) \newline (0, -2) \rarr (-2, 2) \newline (0, -1) \rarr (-1, 1)

If we plot our trapezodial, we find that it is a type II region. The left curve being $u = -v$ and the right curve $u = v$ .

Which means our integral becomes:

\dfrac{1}{2} \int_1^2 \int_{-v}^{v} e^{\dfrac{u}{v}}\ du\ dv

\dfrac{1}{2} \int_1^2 ve^{\dfrac{u}{v}} \bigg\rvert_{u = -v}^{u = v} \ dv

\dfrac{1}{2} e - e^{-1} \int_1^2 v \ dv

\dfrac{1}{2} e - e^{-1} \dfrac{v^2}{2} \bigg\rvert_{v = 1}^{v = 2}

\boxed{\dfrac{3(e - e^{-1})}{4}}

In three variables

It is the same approach. Let $x = x(u, v, w), y = y(u, v, w), z = z(u, v, w)$

The jacobian is:

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = det \begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \newline \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \newline \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{bmatrix}

Just as we did before:

\iiint_E f(x, y, z)\ dV = \iiint_S f(x(u, v, w), y(u, v, w), z(u, v, w)) \vert \tfrac{\partial(x, y, z)}{\partial(u, v, w)} \vert \ dV

Let’s now cover some special cases that we’ll often encounter.

Cylindrical coordinates

Is just like polar coordinates but for three variables:

(x, y, z) \rarr (r, \theta, z)

If we have symmetry around the z-axis, it makes sense to use these variables.

We know that:

x = r cos(\theta) \newline y = r sin(\theta) \newline z = z

Therefore, the jacobian for this becomes:

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = det \begin{bmatrix} cos(\theta) & -r sin(\theta) & 0 \newline sin(\theta) & r cos(\theta) & 0 \newline 0 & 0 & 1 \end{bmatrix}

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = 0 \cdot det \begin{bmatrix} -r sin(\theta) & 0 \newline r cos(\theta) & 0 \newline \end{bmatrix} - 0 \cdot det \begin{bmatrix} cos(\theta) & 0 \newline sin(\theta) & 0 \newline \end{bmatrix} + 1 \cdot det \begin{bmatrix} cos(\theta) & -r sin(\theta) \newline sin(\theta) & r cos(\theta) \newline \end{bmatrix}

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = r cos^2(\theta) + r sin^2(\theta) = r(cos^2(\theta) + sin^2(\theta)) = \boxed{r}

Change to spherical coordinates

\iiint_E f(x, y, z)\ dV = \iiint_S f(r cos(\theta), r sin(\theta), z) \cdot r \ dV

Example 2

Find

\iint_E \sqrt{x^2 + y^2}\ dV

Where $E$ is the solid lying within the cylinder, $x^2 + y^2 = 1$ , below the plane $z = 4$ and above the paraboloid, $z = 1 - x^2 - y^2$ .

Let us describe the solid $E$ :

E = \{(x, y, z) | 0 \leq x^2 + y^2 \leq 1, 1 - x^2 - y^2 \leq z \leq 4\}

In spherical coordinates:

E = \{(x, y, z) | 0 \leq r \leq 1, 1 - r^2 \leq z \leq 4\}

Let’s make $z$ our inner variable.

\iint_E \sqrt{x^2 + y^2}\ dV = \iiint_S \sqrt{r^2} \cdot r\ dV

\iiint_S r^2\ dV

\int_0^{2\pi} \int_0^1 \int_{1 - r^2}^{4} r^2\ dz\ dr\ d\theta

\int_0^{2\pi} \int_0^1 r^2 z \bigg\rvert_{z = 1 - r^2}^{z = 4} dr\ d\theta

\int_0^{2\pi} \int_0^1 3r^2 + r^4\ dr\ d\theta

\int_0^{2\pi} \int_0^1 r^3 + \dfrac{r^5}{5} \bigg\rvert_{r = 0}^{r = 1} dr\ d\theta

\int_0^{2\pi} \dfrac{6}{5} d\theta

\dfrac{6}{5}\theta \bigg\rvert_{\theta = 0}^{\theta = 2\pi}

\boxed{\dfrac{12\pi}{5}}

Spherical coordinates

In this coordinate system we introduce a completely new idea.

Let:

\rho - \text{distance form origin} \newline \varphi - \text{angle between positive } z\text{-axis} \newline \theta- \text{angle in } x\text{-plane}\newline

This means:

\rho \geq 0 \newline 0 \leq \varphi \leq \pi \newline 0 \leq \theta \leq 2\pi

r = \rho sin(\varphi) \newline x = r cos(\theta) = \rho sin(\varphi) cos(\theta) \newline y = r sin(\theta) = \rho sin(\varphi) sin(\theta) \newline z = \rho cos(\varphi)

The jacobian:

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = det \begin{bmatrix} sin(\varphi) cos(\theta) & \rho cos(\varphi) cos(\theta) & -\rho sin(\varphi) sin(\theta) \newline sin(\varphi) sin(\theta) & \rho cos(\varphi) sin(\theta) & \rho sin(\varphi) cos(\theta) \newline cos(\varphi) & -\rho sin(\varphi) & 0 \end{bmatrix}

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = \ldots = \rho^2 sin(\varphi)

Changing to spherical coordinates

\iiint_E f(x, y, z)\ dV = \iiint_S f(\rho sin(\varphi) cos(\theta), \rho sin(\varphi) sin(\theta), \rho cos(\varphi)) \cdot \rho^2 sin(\varphi) \ dV

It makes sense to change to spherical coordinates when our solid is bounded by spheres and/or cones.

Example 3

Find

\iint_B e^{(x^2 + y^2 + z^2)^{\frac{3}{2}}}\ dV

Where $B$ is the unit ball.

B = \{(x, y, z) | x^2 + y^2 + z^2 \leq 1 \}

In spherical coordinates:

S = \{(\rho, \varphi, \theta) | 0 \leq \rho \leq 1, 0 \leq \varphi \leq \pi, 0 \leq \theta \leq 2\pi \}

Which is easy to integrate over.

\iint_B e^{(x^2 + y^2 + z^2)^{\frac{3}{2}}}\ dV = \iiint_S e^{(\rho^2)^{\frac{3}{2}}} \cdot \rho^2 sin(\varphi)\ dV

\iiint_S e^{\rho^3} \cdot \rho^2 sin(\varphi)\ dV

\int_0^{2\pi} \int_0^1 \int_0^{\pi} e^{\rho^3} \cdot \rho^2 sin(\varphi)\ d\varphi\ d\rho\ d\theta

\int_0^{2\pi} \int_0^1 e^{\rho^3} \cdot \rho^2 \left[-cos(\varphi) \bigg\rvert_{\varphi = 0}^{\varphi = \pi} \right]\ d\rho\ d\theta

\int_0^{2\pi} \int_0^1 2e^{\rho^3} \cdot \rho^2\ d\rho\ d\theta

Let

u = \rho^3

du = 3 \rho^2 d\rho

\frac{du}{3} = \rho^2 d\rho

\int_0^{2\pi} \int_0^1 \frac{2}{3} e^u\ du\ d\theta

\int_0^{2\pi} \frac{2}{3} e^u \bigg\rvert_{u = 0}^{u = 1} d\theta

\int_0^{2\pi} \frac{2(e - 1)}{3} d\theta

\int_0^{2\pi} \frac{2(e - 1)}{3} d\theta

\frac{2(e - 1)}{3} \theta \bigg\rvert_{\theta = 0}^{\theta = 2\pi}

\boxed{\frac{4\pi(e - 1)}{3}}

Some applications of multiple integrals

Let’s see what we can achieve with multiple integrals.

Computing mass of solid

Let $E$ be an arbitrary solid, with a density function, $\rho(x, y, z)$

We say that the mass of the solid is:

\iiint_E \rho(x, y, z)\ dV

We perform the same proof as for integrals for solids, but with the density function instead. Taking the sum and limit of small points and their densities.

Center of Mass

One very important application is finding the center of mass of a solid.

Given a physical system with $m_i$ points, the center of mass has the coordinates, $(x_0, y_0, z_0)$ .

Therefore:

x_0 = \dfrac{\sum m_i x_i}{\sum m_i}

y_0 = \dfrac{\sum m_i y_i}{\sum m_i}

z_0 = \dfrac{\sum m_i z_i}{\sum m_i}

For an arbitrary solid, it’s almost sum same:

x_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot x\ dV}{\iiint_E \rho(x, y, z)\ dV}

y_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot y\ dV}{\iiint_E \rho(x, y, z)\ dV}

z_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot z\ dV}{\iiint_E \rho(x, y, z)\ dV}

Again, the proof is the exact same as the previous one.

Example 4

A rectangular solid of width 2, length 10 and height 2 has the density function, $\rho(x, y, z) = y^3$ , where $x, y, z$ are counted from the lower left farthest corner. Now the solid is place on a table, such that 2/3 of its length, counting from the left, lies on the table. Does the rectangular solid fall of the table?

So, we need to compute the center of mass of the solid. If we think logically about this, the only coordinate that matters is the $y$ coordinate.

If $y_0 \leq \dfrac{2}{3} \cdot 10$ it stays, otherwise it falls from the table. So let’s compute $y_0$ .

y_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot y\ dV}{m}

y_0 = \dfrac{\iiint_E y^3 \cdot y\ dV}{m}

y_0 = \dfrac{\iiint_E y^4 \ dV}{m}

Since we have a rectangular box:

y_0 = \dfrac{\int_0^2 \int_0^2 \int_0^{10} y^4\ dy\ dx\ dz}{m}

y_0 = \dfrac{\int_0^2 \int_0^2 \dfrac{10^5}{5}\ dx\ dz}{m}

y_0 = \dfrac{\int_0^2 \dfrac{10^5}{5} \cdot 2\ dz}{m}

y_0 = \dfrac{\dfrac{10^5}{5} \cdot 2 \cdot 2}{m}

Now let’s compute $m$ :

m = \iiint_E \rho(x, y, z)\ dV

m = \iiint_E y^3\ dV

m = \int_0^2 \int_0^2 \int_0^{10} y^3\ dy\ dx\ dz

m = \int_0^2 \int_0^2 \dfrac{10^4}{4}dx\ dz

m = \int_0^2 \dfrac{10^4}{4} \cdot 2\ dz

m = \dfrac{10^4}{4} \cdot 2 \cdot 2

y_0 = \dfrac{\dfrac{10^5}{5} \cdot 2 \cdot 2}{\dfrac{10^4}{4} \cdot 2 \cdot 2} = \ldots = \boxed{8}