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Part 13 - The Jacobian

LMA017
September 26, 2023
10 min read
Table of Contents
LMA017_13

Introduction

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

General case of variable change

Df(x,y) dA \iint_D f(x, y)\ dA

We want to rewrite this in terms of uu and vv. Let’s say that: x=g(u,v)y=h(u,v) x = g(u, v) \newline y = h(u, v)

We can say that we have a transformation, TT, from the uvuv-plane to the xyxy-plane, given by our functions gg and hh.

We need to make some assumptions of TT to make our lives easier.

  1. TT is a C1C^1-transformation, meaning that gg and hh have continuous partial derivatives.

  2. TT is an injective transformation (meaning it is 1-to-1). This means that we can express uu and vv in terms of xx and yy.

With this, we can define the Jacobian matrix.

Definition

The jacobian matrix of a transformation, TT, is:

(x,y)(u,v)=det[xuxvyuyv] \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}

Suppose TT is a transformation, from the uvuv-plane to the xyxy-plane. Assuming that the jacobian for TT, is non-zero, then: Df(x,y) dA=Sf(x(u,v),y(u,v))(x,y)(u,v) dA \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

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

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

Let u=x+yu = x + y and v=xyv = x - y.

This means that: x=u+v2y=uv2 x = \frac{u + v}{2} \newline y = \frac{u - v}{2}

Let’s compute the jacobian: (x,y)(u,v)=det[xuxvyuyv] \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}

(x,y)(u,v)=det[12121212] \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}

(x,y)(u,v)=12 \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)(1,1)(2,0)(2,2)(0,2)(2,2)(0,1)(1,1) (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=vu = -v and the right curve u=vu = v.

Which means our integral becomes: 1212vveuv du dv \dfrac{1}{2} \int_1^2 \int_{-v}^{v} e^{\dfrac{u}{v}}\ du\ dv

1212veuvu=vu=v dv \dfrac{1}{2} \int_1^2 ve^{\dfrac{u}{v}} \bigg\rvert_{u = -v}^{u = v} \ dv

12ee112v dv \dfrac{1}{2} e - e^{-1} \int_1^2 v \ dv

12ee1v22v=1v=2 \dfrac{1}{2} e - e^{-1} \dfrac{v^2}{2} \bigg\rvert_{v = 1}^{v = 2}

3(ee1)4 \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)x = x(u, v, w), y = y(u, v, w), z = z(u, v, w)

The jacobian is: (x,y,z)(u,v,w)=det[xuxvxwyuyvywzuzvzw] \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: Ef(x,y,z) dV=Sf(x(u,v,w),y(u,v,w),z(u,v,w))(x,y,z)(u,v,w) dV \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)(r,θ,z) (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=rcos(θ)y=rsin(θ)z=z x = r cos(\theta) \newline y = r sin(\theta) \newline z = z

Therefore, the jacobian for this becomes: (x,y,z)(u,v,w)=det[cos(θ)rsin(θ)0sin(θ)rcos(θ)0001] \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}

(x,y,z)(u,v,w)=0det[rsin(θ)0rcos(θ)0]0det[cos(θ)0sin(θ)0]+1det[cos(θ)rsin(θ)sin(θ)rcos(θ)] \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}

(x,y,z)(u,v,w)=rcos2(θ)+rsin2(θ)=r(cos2(θ)+sin2(θ))=r \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

Ef(x,y,z) dV=Sf(rcos(θ),rsin(θ),z)r dV \iiint_E f(x, y, z)\ dV = \iiint_S f(r cos(\theta), r sin(\theta), z) \cdot r \ dV

Example

Find Ex2+y2 dV \iint_E \sqrt{x^2 + y^2}\ dV

Where EE is the solid lying within the cylinder, x2+y2=1x^2 + y^2 = 1, below the plane z=4z = 4 and above the paraboloid, z=1x2y2z = 1 - x^2 - y^2.

Let us describe the solid EE: E={(x,y,z)0x2+y21,1x2y2z4} 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)0r1,1r2z4} E = \{(x, y, z) | 0 \leq r \leq 1, 1 - r^2 \leq z \leq 4\}

Let’s make zz our inner variable. Ex2+y2 dV=Sr2r dV \iint_E \sqrt{x^2 + y^2}\ dV = \iiint_S \sqrt{r^2} \cdot r\ dV

Sr2 dV \iiint_S r^2\ dV

02π011r24r2 dz dr dθ \int_0^{2\pi} \int_0^1 \int_{1 - r^2}^{4} r^2\ dz\ dr\ d\theta

02π01r2zz=1r2z=4dr dθ \int_0^{2\pi} \int_0^1 r^2 z \bigg\rvert_{z = 1 - r^2}^{z = 4} dr\ d\theta

02π013r2+r4 dr dθ \int_0^{2\pi} \int_0^1 3r^2 + r^4\ dr\ d\theta

02π01r3+r55r=0r=1dr dθ \int_0^{2\pi} \int_0^1 r^3 + \dfrac{r^5}{5} \bigg\rvert_{r = 0}^{r = 1} dr\ d\theta

02π65dθ \int_0^{2\pi} \dfrac{6}{5} d\theta

65θθ=0θ=2π \dfrac{6}{5}\theta \bigg\rvert_{\theta = 0}^{\theta = 2\pi}

12π5 \boxed{\dfrac{12\pi}{5}}

Spherical coordinates

In this coordinate system we introduce a completely new idea.

Let: ρdistance form originφangle between positive z-axisθangle in x-plane \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: ρ00φπ0θ2π \rho \geq 0 \newline 0 \leq \varphi \leq \pi \newline 0 \leq \theta \leq 2\pi

r=ρsin(φ)x=rcos(θ)=ρsin(φ)cos(θ)y=rsin(θ)=ρsin(φ)sin(θ)z=ρcos(φ) 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: (x,y,z)(u,v,w)=det[sin(φ)cos(θ)ρcos(φ)cos(θ)ρsin(φ)sin(θ)sin(φ)sin(θ)ρcos(φ)sin(θ)ρsin(φ)cos(θ)cos(φ)ρsin(φ)0] \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}

(x,y,z)(u,v,w)==ρ2sin(φ) \dfrac{\partial(x, y, z)}{\partial(u, v, w)} = \ldots = \rho^2 sin(\varphi)

Changing to spherical coordinates

Ef(x,y,z) dV=Sf(ρsin(φ)cos(θ),ρsin(φ)sin(θ),ρcos(φ))ρ2sin(φ) dV \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

Find Be(x2+y2+z2)32 dV \iint_B e^{(x^2 + y^2 + z^2)^{\frac{3}{2}}}\ dV

Where BB is the unit ball.

B={(x,y,z)x2+y2+z21} B = \{(x, y, z) | x^2 + y^2 + z^2 \leq 1 \}

In spherical coordinates: S={(ρ,φ,θ)0ρ1,0φπ,0θ2π} 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. Be(x2+y2+z2)32 dV=Se(ρ2)32ρ2sin(φ) dV \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

Seρ3ρ2sin(φ) dV \iiint_S e^{\rho^3} \cdot \rho^2 sin(\varphi)\ dV

02π010πeρ3ρ2sin(φ) dφ dρ dθ \int_0^{2\pi} \int_0^1 \int_0^{\pi} e^{\rho^3} \cdot \rho^2 sin(\varphi)\ d\varphi\ d\rho\ d\theta

02π01eρ3ρ2[cos(φ)φ=0φ=π] dρ dθ \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

02π012eρ3ρ2 dρ dθ \int_0^{2\pi} \int_0^1 2e^{\rho^3} \cdot \rho^2\ d\rho\ d\theta

Let u=ρ3 u = \rho^3

du=3ρ2dρ du = 3 \rho^2 d\rho

du3=ρ2dρ \frac{du}{3} = \rho^2 d\rho

02π0123eu du dθ \int_0^{2\pi} \int_0^1 \frac{2}{3} e^u\ du\ d\theta

02π23euu=0u=1dθ \int_0^{2\pi} \frac{2}{3} e^u \bigg\rvert_{u = 0}^{u = 1} d\theta

02π2(e1)3dθ \int_0^{2\pi} \frac{2(e - 1)}{3} d\theta

02π2(e1)3dθ \int_0^{2\pi} \frac{2(e - 1)}{3} d\theta

2(e1)3θθ=0θ=2π \frac{2(e - 1)}{3} \theta \bigg\rvert_{\theta = 0}^{\theta = 2\pi}

4π(e1)3 \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 EE be an arbitrary solid, with a density function, ρ(x,y,z)\rho(x, y, z)

We say that the mass of the solid is: Eρ(x,y,z) dV \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 mim_i points, the center of mass has the coordinates, (x0,y0,z0)(x_0, y_0, z_0).

Therefore: x0=miximi x_0 = \dfrac{\sum m_i x_i}{\sum m_i}

y0=miyimi y_0 = \dfrac{\sum m_i y_i}{\sum m_i}

z0=mizimi z_0 = \dfrac{\sum m_i z_i}{\sum m_i}

For an arbitrary solid, it’s almost sum same: x0=Eρ(x,y,z)x dVEρ(x,y,z) dV x_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot x\ dV}{\iiint_E \rho(x, y, z)\ dV}

y0=Eρ(x,y,z)y dVEρ(x,y,z) dV y_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot y\ dV}{\iiint_E \rho(x, y, z)\ dV}

z0=Eρ(x,y,z)z dVEρ(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

A rectangular solid of width 2, length 10 and height 2 has the density function, ρ(x,y,z)=y3\rho(x, y, z) = y^3, where x,y,zx, 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 yy coordinate.

If y02310y_0 \leq \dfrac{2}{3} \cdot 10 it stays, otherwise it falls from the table. So let’s compute y0y_0.

y0=Eρ(x,y,z)y dVm y_0 = \dfrac{\iiint_E \rho(x, y, z) \cdot y\ dV}{m}

y0=Ey3y dVm y_0 = \dfrac{\iiint_E y^3 \cdot y\ dV}{m}

y0=Ey4 dVm y_0 = \dfrac{\iiint_E y^4 \ dV}{m}

Since we have a rectangular box: y0=0202010y4 dy dx dzm y_0 = \dfrac{\int_0^2 \int_0^2 \int_0^{10} y^4\ dy\ dx\ dz}{m}

y0=02021055 dx dzm y_0 = \dfrac{\int_0^2 \int_0^2 \dfrac{10^5}{5}\ dx\ dz}{m}

y0=0210552 dzm y_0 = \dfrac{\int_0^2 \dfrac{10^5}{5} \cdot 2\ dz}{m}

y0=105522m y_0 = \dfrac{\dfrac{10^5}{5} \cdot 2 \cdot 2}{m}

Now let’s compute mm: m=Eρ(x,y,z) dV m = \iiint_E \rho(x, y, z)\ dV

m=Ey3 dV m = \iiint_E y^3\ dV

m=0202010y3 dy dx dz m = \int_0^2 \int_0^2 \int_0^{10} y^3\ dy\ dx\ dz

m=02021044dx dz m = \int_0^2 \int_0^2 \dfrac{10^4}{4}dx\ dz

m=0210442 dz m = \int_0^2 \dfrac{10^4}{4} \cdot 2\ dz

m=104422 m = \dfrac{10^4}{4} \cdot 2 \cdot 2

y0=105522104422==8 y_0 = \dfrac{\dfrac{10^5}{5} \cdot 2 \cdot 2}{\dfrac{10^4}{4} \cdot 2 \cdot 2} = \ldots = \boxed{8}