# Calculus/Change of variables

 ← Multiple integration Calculus Vector calculus → Change of variables

The Jacobian matrix and the change of variables are proven to be extremely useful in multivariable calculus when we want to change our variables. They are extremely useful because if we want to integrate a function such as

$\iint _{R}e^{\frac {x+y}{x-y}}\,dA$ , where $R$ is the trapezoidal region with vertices $(1,0),(2,0),(0,-2),(0,-1)$ ,

it would be helpful if we can substitute $x+y$ as $u$ and $x-y$ as $v$ because $e^{\frac {u}{v}}$ is easier to be integrated. However, we need to be familiar with integration, transformation, and the Jacobian, which the latter two will be discussed in this chapter.

## Transformation

Let us start with an introduction to the process of variable transformation. Assume that we have a function $f(x,y)$ . We want to calculate the expression:

$\iint _{R}f(x,y)\,dA$

in which $R$  is a region in $xy$ -plane. (Another notation for $dA$  is $dxdy$ . ($dA$  here is not differential.) )

However, the area of $R$  is too complicated to be written out in terms of $x,y$ . So, we want to change the variables so that the area of $R$  can be more easily expressed. Furthermore, the function itself is too hard to be integrated.

It would be much easier if the variables can be changed to more convenient ones, Assume there are two more variables $u,v$  that have connections with variables $x,y$  that satisfy:

$x=x(u,v)\quad {\text{and}}\quad y=y(u,v).$

The original integral can be rewritten into:

$\iint _{S}f(x(u,v),y(u,v))\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\,du\,dv$

in which $S$  is another region in $uv$ -plane transformed from the region $R$  in $xy$ -plane. The purpose of this section is to have us understand the process of this transformation, excluding the $\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|$  part. We will discuss the purpose and meaning of $\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|$  in the next section.

### Introduction

In fact, we have already encountered two examples of variable transformation in $\mathbb {R} ^{3}$ .

The first example is using polar coordinates in integration while the second one is using spherical coordinates in integration. Using polar coordinates in integration is a change in variable because we effectively change the variables $x,y,z$  into $r,\theta ,z$  with relations:

$x=r\cos \theta \quad y=r\sin \theta \quad z=z$

As a result, the function being integrated $f(x,y)$  is transformed into $f(x(r,\theta ,z),y(r,\theta ,z),z(r,\theta ,z))$ , thus giving us:

$\iiint _{R}f(x,y,z)dV=\iiint _{S}f(r\cos \theta ,r\sin \theta ,z)dz\ r\ dr\ d\theta$ , which is the formula for polar coordinates integration. (It will be proved later)

The second example, integration in spherical coordinates, offers a similar explanation. The original variables $x,y,z$  and the transformed variables $\rho ,\theta ,\phi$  have the relations:

${\begin{cases}x=\rho \sin \phi \cos \theta \\y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi \end{cases}}$

These relations can give us that

$\iiint _{R}f(x,y,z)dV=\iiint _{S}f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi )\rho ^{2}\sin \phi \ d\rho \ d\theta \ d\phi$ , which is the formula for spherical coordinates integration. (It will be proved later)

### Generalization

We understand the transformation from Cartesian coordinates to both polar and spherical coordinates. However, those two are specific examples of variable transformation. We should expand our scope into all kinds of transformation. Instead of specific changes, such as $x=r\cos \theta ,y=r\sin \theta {\text{ and }}z=z$ , we will talk about general changes. Let's start from two variables.

We consider a change of variables that is given by a transformation $T$  from the $uv$ -plane to the $xy$ -plane. In other words,

$T(u,v)=(x,y)$ , where $(x,y)$  is the original or old variables and $(u,v)$  is the new ones.

In this transformation, $x,y$  are related to $u,v$  by the equations

$x=g(u,v)\quad y=h(u,v)$

We usually just assume that $T$  is a $C^{1}$  transformation, which means that $g,h$  have continuous 1st-order partial derivatives. Now, time for some terminologies.

• If $T(u_{1},v_{1})=(x_{1},y_{1})$ , the point $(x_{1},y_{1})$  is called the image of the point $(u_{1},v_{1})$ .
• If no two points have the same image, like functions, $T$ , the transformation, is called one-to-one (or injective).
• $T$  transforms region $S$  into region $R$ . $R$  is called the image of $S$ . The transformation can be described as:

$T(S)=R$

• If $T$  is one-to-one, then, like functions, it has an inverse transformation $T^{-1}$  from the $xy$ -plane to the $uv$ -plane, with relation

$T^{-1}(R)=S$

### Regions

Recall that we have established the transformation $T(S)=R$ , where $S$  is the region in the $uv$ -plane while $R$  is the region in the $xy$ -plane. If we are given the region $S$  and transformation $T$ , we are expected to calculate the region $R$ . For example, a transformation is defined by the equations

$x=u^{2}-v^{2}\quad y=2uv$

Find the image of $S$ , which is defined as $S=\{(u,v):0\leq u\leq 1,\ 0\leq v\leq 1\}$ .

In this case, we need to know the boundaries of the region $S$ , which is confined by the lines:

$u=0\quad u=1\quad v=0\quad v=1$

If we can redefine the boundaries using $x,y$  instead of $u,v$ , we effectively will find the image of $S$ .

{\begin{aligned}{\text{When }}\quad &u=0\quad (0\leq v\leq 1)\\&{\begin{cases}x=-v^{2}\\y=0\\\end{cases}}\quad {\text{(substitution)}}\\{\text{Thus, }}\quad &y=0\quad (-1\leq x\leq 0)\\\end{aligned}} {\begin{aligned}{\text{When }}\quad &u=1\quad (0\leq v\leq 1)\\&{\begin{cases}x=1-v^{2}\\y=2v\\\end{cases}}\\{\text{Thus, }}\quad &x=1-{\frac {y^{2}}{4}}\quad (0\leq x\leq 1)\\\end{aligned}} {\begin{aligned}{\text{When }}\quad &v=0\quad (0\leq 0\leq 1)\\&{\begin{cases}x=u^{2}\\y=0\\\end{cases}}\\{\text{Thus, }}\quad &y=0\quad (0\leq x\leq 1)\\\end{aligned}} {\begin{aligned}{\text{When }}\quad &v=1\quad (0\leq u\leq 1)\\&{\begin{cases}x=u^{2}-1\\y=2u\\\end{cases}}\\{\text{Thus, }}\quad &x={\frac {y^{2}}{4}}-1\quad (-1\leq x\leq 0)\\\end{aligned}}

As a result, the image of $S$  is $R=\{(x,y):0\leq y\leq 2,\ {\frac {y^{2}}{4}}-1\leq x\leq 1-{\frac {y^{2}}{4}}\}$

We can use the same method to calculate $S$  from $R$ .

## The Jacobian

The Jacobian matrix is one of the most important concept in this chapter. It "compromises" the change in area when we change the variables so that after changing the variables, the result of the integral does not change. Recall that at the very beginning of the last section, we reserved the explanation of $\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|$  from $\iint _{S}f(x(u,v),y(u,v))\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\ du\ dv$  here. To actually start explaining that, we should review some basic concepts.

### Review "u-substitution"

Recall that when we are discussing $u$ -substitution (a simple way to describe "integration by substitution for single-variable functions"), we use the following method to solve integrals.

$\int _{a}^{b}f(x)\ dx=\int _{c}^{d}f(x(u))\ {\frac {dx}{du}}\ du\quad {\text{where }}c=x(a),d=x(b)$

For example,

$\int {\frac {\sin(\ln(x))}{x}}\ dx$

{\begin{aligned}{\text{Let }}\quad &u=\ln(x)\\{\text{Thus, }}\quad &{\frac {du}{dx}}={\frac {1}{x}}\\\Rightarrow \ &du={\frac {1}{x}}dx\\\end{aligned}}

{\begin{aligned}\int {\frac {\sin(\ln(x))}{x}}\ dx&=\int \sin(\ln(x))\ {\Big (}{\frac {1}{x}}\ dx{\Big )}&\quad {\text{rearrangement}}\\&=\int \sin(u)\ du&\quad {\text{let }}u=\ln(x)\implies du={\frac {1}{x}}\ dx\\&=-\cos(u)+C&\quad {\text{integration}}\\&=-\cos(\ln(x))+C&\quad {\text{resubstitution}}\\\end{aligned}}

If we add endpoints into the integral, the result will be:

{\begin{aligned}\int _{e}^{e^{2}}{\frac {\sin(\ln(x))}{x}}\ dx&=\int _{e}^{e^{2}}\sin(\ln(x))\ {\Big (}{\frac {1}{x}}\ dx{\Big )}&\quad {\text{rearrangement}}\\&=\int _{1}^{2}\sin(u)\ du&\quad {\text{remember }}u=\ln(x){\text{ and }}du={\frac {1}{x}}\ dx\\&={\Big [}-\cos(u){\Big ]}_{1}^{2}&\quad {\text{integration}}\\&=\cos(1)-\cos(2)\\\end{aligned}}

If we look carefully at the "rearrangement" and "remember" part in the solution, we find that we effectively changed our variable from $x$  to $u$  through this method:

$\int _{a}^{b}f(x)dx=\int _{x=a}^{x=b}f(x(u))\ d(x(u))=\int _{u=x(a)}^{u=x(b)}f(x(u))\ {\frac {dx}{du}}\ du$ , which is what we have mentioned above.

The appearance of the term ${\frac {dx}{du}}$  not only is a mathematical product of deduction, but also serves a intuitive purpose. When we change our function from $f(x)$  to $f(x(u))$ , we also change the region we are integrating, which can be seen by looking at the endpoints. This change of region is either "stretched" or "condensed" by a factor of ${\frac {du}{dx}}$ . To counter this change, ${\frac {dx}{du}}$  is deduced to compromise (recall that ${\frac {dx}{du}}\left({\frac {du}{dx}}\right)=1$ ). We can simply think this term as a compromise factor that counters the change of region due to a change of variables.

Now, let us put our focus back to two variables. If we change our variables from $x,y$  to $u,v$ , we also change the region we are integrating, as demonstrated in the previous section.

So, continuing our flow of thought, there should also be a term deduced to counter the change of region. In other words:

$\iint _{R}f(x,y)\,dA_{1}=\iint _{S}f(x(u,v),y(u,v))\underbrace {\frac {dA_{1}}{dA_{2}}} _{\text{informal term}}\,dA_{2}$

Note that the symbols used here are for intuitive purpose and not for official use. Official terms will be introduced later in the chapter, but for now, we use these terms for better understanding.

In this case, when we change the function from $f(x,y)$  to $f(x(u),y(u))$ , we "stretched" or "condensed" the area of our region, by a factor of ${\frac {dA_{2}}{dA_{1}}}$ ; therefore, we need to counter the change with a factor of ${\frac {dA_{1}}{dA_{2}}}$ . The Jacobian matrix for two variables is basically an expression for calculating ${\frac {dA_{1}}{dA_{2}}}$  in terms of $u,v$ , so that we are able to integrate the new integral after transformation, since the function involved in the new integral can only in terms of $u,v$ , but not $x,y$  (we need to express $x$  and $y$  in terms of $u,v$ ).

### The Jacobian

#### Double integrals

Now, it is time for us to deduce the Jacobian matrix. In the review above, we already established informally that the Jacobian matrix for two variables is basically ${\frac {dA_{1}}{dA_{2}}}$ , with $dA_{1}$  being the infinitesimally small area in the region $R$  in the $xy$ -plane and $dA_{2}$  being the infinitesimally small area in the region $S$  in the $uv$ -plane. Since we are changing our variables from $x,y$  to $u,v$ , we should describe $dA_{1}$  and $dA_{2}$  in terms of $u,v$  over a region in $uv$ -plane.

Let us start with $dA_{2}$  first because it is easier to calculate. We start with a small rectangle $S_{0}$ , which is a part of $S$ , in the $uv$ -plane whose lower left corner is the point $(u_{0},v_{0})$  and whose dimensions are $\Delta u,\Delta v$ . Thus, the area of $S_{0}$  is

$\Delta A_{2}=\Delta u\Delta v$

The image of $S_{0}$ , in this case let's name it $R_{0}$ , is in the $xy$ -plane according to the transformation $T(S_{0})=R_{0}$ . One of its boundary points is $(x_{0},y_{0})=T(u_{0},v_{0})$ . We can use a vector $\mathbf {r}$  to describe the position vector of $R_{0}$  of the point $(u,v)$ . In other words, $\mathbf {r}$  can describe the region $R_{0}$  given that

$\mathbf {r} (u,v)=x(u,v)\mathbf {i} +y(u,v)\mathbf {j} \quad {\text{where }}u_{0}\leq u\leq u_{0}+\Delta u,\ v_{0}\leq v\leq v_{0}+\Delta v$

The region $R_{0}$  now can be described in terms of $u,v$ . The next step is to utilize the position vector $\mathbf {r} (u,v)$  to calculate its area $dA_{1}$ .

The shape of the region $R_{0}$  after transformation $R_{0}=T(S_{0})$  can be approximated, which is a parallelogram. As we learnt in algebra, the area of a parallelogram is defined to be the product of its base and height. However, this definition cannot help us with our calculations. Instead, we will use the cross product to determine its area. Recall that the area of a parallelogram formed by vectors $\mathbf {a}$  and $\mathbf {b}$  can be calculated by taking the magnitude of the cross product of the two vectors.

$\Delta A_{1}=|\mathbf {a} \times \mathbf {b} |$

In this parallelogram, the two vectors $\mathbf {a}$  and $\mathbf {b}$  are, in terms of $u,v$ :

$\mathbf {a} =\mathbf {r} (u_{0}+\Delta u,v_{0})-\mathbf {r} (u_{0},v_{0})\quad {\text{ and }}\quad \mathbf {b} =\mathbf {r} (u_{0},v_{0}+\Delta v)-\mathbf {r} (u_{0},v_{0})$

It seems very similar to the definition of partial derivatives:

$\mathbf {r} _{u}=\lim _{\Delta u\rightarrow 0}{\frac {\mathbf {r} (u_{0}+\Delta u,v_{0})-\mathbf {r} (u_{0},v_{0})}{\Delta u}}\quad {\text{ and }}\quad \mathbf {r} _{v}=\lim _{\Delta v\rightarrow 0}{\frac {\mathbf {r} (u_{0},v_{0}+\Delta v)-\mathbf {r} (u_{0},v_{0})}{\Delta v}}$

As a result, we can approximate that:

{\begin{aligned}&\mathbf {a} =\mathbf {r} (u_{0}+\Delta u,v_{0})-\mathbf {r} (u_{0},v_{0})\approx \Delta u\mathbf {r} _{u}\quad {\text{ and }}\quad \\&\mathbf {b} =\mathbf {r} (u_{0},v_{0}+\Delta v)-\mathbf {r} (u_{0},v_{0})\approx \Delta v\mathbf {r} _{v}\\\end{aligned}}

Now, we calculate $\mathbf {r} _{u},\mathbf {r} _{v}$ , given that $\mathbf {r} (u,v)=x(u,v)\mathbf {i} +y(u,v)\mathbf {j}$ :

{\begin{aligned}&\mathbf {r} _{u}=x_{u}(u,v)\ \mathbf {i} +y_{u}(u,v)\ \mathbf {j} ={\frac {\partial x}{\partial u}}\ \mathbf {i} +{\frac {\partial y}{\partial u}}\ \mathbf {j} \quad {\text{ and }}\\&\mathbf {r} _{v}=x_{v}(u,v)\ \mathbf {i} +y_{v}(u,v)\ \mathbf {j} ={\frac {\partial x}{\partial v}}\ \mathbf {i} +{\frac {\partial y}{\partial v}}\ \mathbf {j} \\\end{aligned}}

We can calculate $\Delta A_{1}=||\mathbf {a} \times \mathbf {b} ||$  (we take absolute value to prevent negative area). You can review the cross product in Chapter 7.1. Note that the inner bar of || is for calculating the magnitude (or norm) while the outer bar of || is for taking the absolute value.

{\begin{aligned}\Delta A_{1}&=||\mathbf {a} \times \mathbf {b} ||\\&=||(\Delta u\ \mathbf {r} _{u})\times (\Delta v\ \mathbf {r} _{v})||&\quad {\text{approximation}}\\&=||\mathbf {r} _{u}\times \mathbf {r} _{v}||\Delta u\Delta v\\&={\begin{vmatrix}{\begin{vmatrix}\det {\begin{pmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\{\frac {\partial x}{\partial u}}&{\frac {\partial y}{\partial u}}&0\\{\frac {\partial x}{\partial v}}&{\frac {\partial y}{\partial v}}&0\\\end{pmatrix}}\\\end{vmatrix}}\end{vmatrix}}\Delta u\Delta v&\quad {\text{cross product}}\\&={\begin{vmatrix}{\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\mathbf {k} \end{vmatrix}}\end{vmatrix}}\Delta u\Delta v&\quad {\text{evaluation}}\\&={\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\underbrace {|\mathbf {k} |} _{1}\end{vmatrix}}\Delta u\Delta v\\\end{aligned}}

Then, we can substitute our newly deduced terms.

${\frac {\Delta A_{1}}{\Delta A_{2}}}={\frac {{\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\end{vmatrix}}\Delta u\Delta v}{\Delta u\Delta v}}={\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\end{vmatrix}}=\left|{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial v}}-{\frac {\partial x}{\partial v}}{\frac {\partial y}{\partial u}}\right|$

Finally, we derived the absolute value of Jacobian. The definition of Jacobian is as follows:

Definition. (The Jacobian for two variables) The Jacobian of the transformation $T$  given by $x=g(u,v)$  and $y=h(u,v)$  whose partial derivatives exist and are continuous, is

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

We will then use the Jacobian in the change of variables in integrals. The absolute value is added to prevent a negative area.

{\begin{aligned}\iint _{R}f(x,y)\,dA&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}f(x_{i},y_{j})\Delta A\\&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}f(x(u_{i},v_{j}),y(u_{i},v_{j}))\Delta A_{2}\\{\text{Since }}&\Delta A_{2}\approx \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\Delta u\Delta v\\\sum _{i=1}^{m}\sum _{j=1}^{n}f(x(u_{i},v_{j}),y(u_{i},v_{j}))\Delta A_{2}&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}f(x(u_{i},v_{j}),y(u_{i},v_{j}))\ \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\Delta u\Delta v\\&\approx \iint _{S}f(x(u,v),y(u,v))\ \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\ du\ dv\\\end{aligned}}

Here is the theorem for the change of variables in a double integral and we have explained intuitively why and how it works, but the above explanations are not proof of this theorem. In particular, we make some approximation, while the statement in the following theorem is equality, and not approximation. The actual proof is quite complicated and advanced, and thus not included here.

Theorem. (Change of variables for double integration) Suppose $T$  is a $C^{1}$  transformation whose Jacobian is nonzero and that maps a region $S$  in the $uv$ -plane onto a region $R$  in the $xy$ -plane injectively, via the change of variables $x=x(u,v)$  and $y=y(u,v)$ . Suppose that $f$  is continuous on $R$ , we have

$\iint _{R}f(x,y)\,dx\,dy=\iint _{S}f(x(u,v),y(u,v))\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\,du\,dv$

Remark.

If we change some notations, we can get

$\iint _{S}f(u(x,y),v(x,y))\left|{\frac {\partial (u,v)}{\partial (x,y)}}\right|\,dx\,dy=\iint _{R}f(u,v)\,du\,dv,$

($T$  maps a region $S$  in $xy$ -plane onto a region $R$  in $uv$ -plane in this case.) which may be a more convenient form to be used sometimes.

Example.

1 Choose correct expression(s) for the Jacobian ${\frac {\partial (u,v)}{\partial (x,y)}}$  in which $x=2u$  and $y=5v$ .

 ${\frac {1}{10}}$ ${\begin{vmatrix}0&{\frac {1}{2}}\\{\frac {1}{5}}&0\end{vmatrix}}$ ${\begin{vmatrix}2&0\\0&5\end{vmatrix}}$ ${\begin{vmatrix}0&2\\5&0\end{vmatrix}}$ $10$ 2 Choose correct expression(s) for the integral $\iint _{D}(x+y)(5x+2y)$  in which $D$  is a region bounded by $y=1-x,x=2-y,x={\frac {-5y}{3}}$  and $y={\frac {1-5x}{2}}$ .

 $3\int _{1}^{2}\int _{0}^{1}uv\,du\,dv$ $-3\int _{0}^{1}\int _{1}^{2}uv\,du\,dv$ $-{\frac {1}{3}}\int _{0}^{1}\int _{1}^{2}uv\,du\,dv$ ${\frac {1}{3}}\int _{1}^{2}\int _{0}^{1}uv\,du\,dv$ ${\frac {1}{3}}\int _{0}^{1}\int _{1}^{2}uv\,du\,dv$ 3 Choose correct statement(s) from the following statements.

 If both $u$ and $v$ are independent from both $x$ and $y$ , the Jacobian ${\frac {\partial (u,v)}{\partial (x,y)}}=0$ . If both $u$ and $v$ are independent from both $x$ and $y$ , the Jacobian ${\frac {\partial (x,y)}{\partial (u,v)}}=0$ . If $u$ is independent from both $x$ and $y$ , while $v$ is dependent from both $x$ and $y$ , the Jacobian ${\frac {\partial (u,v)}{\partial (x,y)}}=0.$ If $u$ is independent from both $x$ and $y$ , while $v$ is dependent from both $x$ and $y$ , the Jacobian ${\frac {\partial (x,y)}{\partial (u,v)}}=0.$ If $v=ku$ in which $k$ is a real number, then the Jacobian ${\frac {\partial (u,v)}{\partial (x,y)}}=k.$ Example.

Consider a region $D$  that is bounded by the lines $y=2-x$ , $y=3-x$ , $y=4x$  and $y=5x$ . Prove that

$\iint _{D}{\frac {y(x+y)}{x}}dxdy={\frac {19}{3}}\left(\ln \left({\frac {6}{5}}\right)+{\frac {1}{30}}\right).$

Proof.

Let $u=x+y$  and $v={\frac {y}{x}}$ , and $D'$  be the transformed region via these changes of variables. Solving these two equations,

${\begin{cases}u=x+y\\v={\frac {y}{x}}\end{cases}}\implies {\begin{cases}y=u-x\\v={\frac {y}{x}}\end{cases}}\implies v={\frac {u-x}{x}}\implies x={\frac {u}{v+1}}\implies y=u-{\frac {u}{v+1}}={\frac {uv}{v+1}}.$

Therefore, the Jacobian for this transformation is
${\frac {\partial (x,y)}{\partial (u,v)}}={\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\end{vmatrix}}={\begin{vmatrix}{\frac {1}{v+1}}&{\frac {-u}{(v+1)^{2}}}\\{\frac {v}{v+1}}&{\frac {(v+1)u-uv}{(v+1)^{2}}}\end{vmatrix}}={\frac {1}{v+1}}\cdot {\frac {u}{(v+1)^{2}}}-{\frac {-u}{(v+1)^{2}}}\cdot {\frac {v}{v+1}}={\frac {u+uv}{(v+1)^{3}}}={\frac {u}{(v+1)^{2}}}.$

Also, the bounds for $x+y$  and ${\frac {y}{x}}$  in $D$  are $2\leq x+y\leq 3$  and $4\leq {\frac {y}{x}}\leq 5$ . So, the bounds for $u$  and $v$  in $D'$  are $2\leq u\leq 3$  and $4\leq v\leq 5$ . Thus, the desired integral is
{\begin{aligned}\iint _{D}{\frac {y(x+y)}{x}}dx&=\int _{4}^{5}\int _{2}^{3}\left(uv\cdot \underbrace {\frac {u}{(v+1)^{2}}} _{>0{\text{ because of the bounds}}}\right)\,du\,dv\\&=\int _{4}^{5}{\frac {v}{(v+1)^{2}}}\left({\frac {3^{3}}{3}}-{\frac {2^{3}}{3}}\right)\,dv\\&={\frac {19}{3}}\int _{4+1}^{5+1}{\frac {w-1}{w^{2}}}\,dw\qquad {\text{let }}w=v+1\implies dw=dv\\&={\frac {19}{3}}\left[\ln |w|-{\frac {1}{w}}\right]_{5}^{6}\\&={\frac {19}{3}}\left(\ln 6-\ln 5+{\frac {1}{30}}\right)\\&={\frac {19}{3}}\left(\ln \left({\frac {6}{5}}\right)+{\frac {1}{30}}\right)\qquad {\text{by properties of logarithm}}\end{aligned}}

#### Triple integrals

If we continue our flow of thoughts, we can also find the Jacobian for three variables. Suppose there is a function $f(x,y,z)$ . $x,y,z$  has relations with $u,v,w$ , which are

$x=x(u,v,w),\quad y=y(u,v,w),\quad {\text{and}}\quad z=z(u,v,w)$

$R$  is a region in the $xyz$ -space, and $S$  is a region in the $uvw$ -space, with transformation $T(S)=R$ .

To calculate the Jacobian for three variables, we go through a similar process. The process of transformation will be: a rectangular prism with dimensions $\Delta u,\Delta v,\Delta w$  in the $uvw$ -space to a parallelepiped in the $xyz$ -space and a volume of $\Delta V_{2}=\Delta u\Delta v\Delta w$ . The parallelepiped can be described with the position vector:

$\mathbf {r} (u,v,w)=x(u,v,w)\ \mathbf {i} +y(u,v,w)\ \mathbf {j} +z(u,v,w)\ \mathbf {k}$

The three sides of the parallelepiped can be described by the position vector as:

{\begin{aligned}&\mathbf {a} =\mathbf {r} (u+\Delta u,v,w)-\mathbf {r} (u,v,w),\\&\mathbf {b} =\mathbf {r} (u,v+\Delta v,w)-\mathbf {r} (u,v,w),\quad {\text{and}}\\&\mathbf {c} =\mathbf {r} (u,v,w+\Delta w)-\mathbf {r} (u,v,w).\\\end{aligned}}

Since the derivatives of $\mathbf {r}$  are defined as:

{\begin{aligned}&\mathbf {r} _{u}=\lim _{\Delta u\rightarrow 0}{\frac {\mathbf {r} (u+\Delta u,v,w)-\mathbf {r} (u,v,w)}{\Delta u}},\\&\mathbf {r} _{v}=\lim _{\Delta v\rightarrow 0}{\frac {\mathbf {r} (u,v+\Delta v,w)-\mathbf {r} (u,v,w)}{\Delta v}},\quad {\text{and}}\\&\mathbf {r} _{w}=\lim _{\Delta w\rightarrow 0}{\frac {\mathbf {r} (u,v,w+\Delta w)-\mathbf {r} (u,v,w)}{\Delta w}}.\\\end{aligned}}

The three vectors $\mathbf {a} ,\mathbf {b} ,\mathbf {c}$  can be similarly approximated into:

$\mathbf {a} \approx \Delta u\ \mathbf {r} _{u},\quad \mathbf {b} \approx \Delta v\ \mathbf {r} _{v},\quad {\text{and}}\quad \mathbf {c} \approx \Delta w\ \mathbf {r} _{w}$

Since the position vector $\mathbf {r}$  is $\mathbf {r} (u,v,w)=x(u,v,w)\ \mathbf {i} +y(u,v,w)\ \mathbf {j} +z(u,v,w)\ \mathbf {k}$ , the partial derivatives for $\mathbf {r}$  are:

{\begin{aligned}&\mathbf {r} _{u}={\frac {\partial x}{\partial u}}\ \mathbf {i} +{\frac {\partial y}{\partial u}}\ \mathbf {j} +{\frac {\partial z}{\partial u}}\ \mathbf {k} ,\\&\mathbf {r} _{v}={\frac {\partial x}{\partial v}}\ \mathbf {i} +{\frac {\partial y}{\partial v}}\ \mathbf {j} +{\frac {\partial z}{\partial v}}\ \mathbf {k} ,\quad {\text{and}}\\&\mathbf {r} _{v}={\frac {\partial x}{\partial w}}\ \mathbf {i} +{\frac {\partial y}{\partial w}}\ \mathbf {j} +{\frac {\partial z}{\partial w}}\ \mathbf {k} \\\end{aligned}}

Recall that the volume of a parallelepiped determined by the vectors $\mathbf {a} ,\mathbf {b} ,\mathbf {c}$  is the magnitude of their scalar triple product:

$V=|(\mathbf {a} \times \mathbf {b} )\ \cdot \ \mathbf {c} |$

We just need to substitute the vectors with what we have yielded.

{\begin{aligned}\Delta V_{1}=|(\mathbf {a} \times \mathbf {b} )\ \cdot \ \mathbf {c} |&=|(\Delta u\ \mathbf {r} _{u})\times (\Delta v\ \mathbf {r} _{v})\ \cdot \ \Delta w\ \mathbf {r} _{w}|\\&=|\mathbf {r} _{u}\times \mathbf {r} _{v}\ \cdot \ \mathbf {r} _{w}|\Delta u\Delta v\Delta w\\&={\begin{vmatrix}{\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\{\frac {\partial x}{\partial u}}&{\frac {\partial y}{\partial u}}&{\frac {\partial z}{\partial u}}\\{\frac {\partial x}{\partial v}}&{\frac {\partial y}{\partial v}}&{\frac {\partial z}{\partial v}}\\\end{vmatrix}}\ \cdot \ {\begin{pmatrix}{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial w}}\\\end{pmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w&\quad {\text{cross product}}\\&={\begin{vmatrix}{\begin{pmatrix}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial v}}-{\frac {\partial z}{\partial u}}{\frac {\partial y}{\partial v}}\\{\frac {\partial z}{\partial u}}{\frac {\partial x}{\partial v}}-{\frac {\partial x}{\partial u}}{\frac {\partial z}{\partial v}}\\{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial v}}-{\frac {\partial y}{\partial u}}{\frac {\partial x}{\partial v}}\\\end{pmatrix}}\ \cdot \ {\begin{pmatrix}{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial w}}\\\end{pmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w\\&=\left|{\frac {\partial x}{\partial w}}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial v}}-{\frac {\partial x}{\partial w}}{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial u}}+{\frac {\partial x}{\partial v}}{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial u}}-{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial v}}+{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial w}}-{\frac {\partial x}{\partial v}}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial w}}\right|\Delta u\Delta v\Delta w&\quad {\text{dot product}}\\&=\left|{\frac {\partial x}{\partial u}}{\bigg (}{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial w}}-{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial v}}{\bigg )}+{\frac {\partial x}{\partial v}}{\bigg (}{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial u}}-{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial w}}{\bigg )}+{\frac {\partial x}{\partial w}}{\bigg (}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial v}}-{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial u}}{\bigg )}\right|\Delta u\Delta v\Delta w&\quad {\text{rearrangement}}\\&={\begin{vmatrix}{\begin{vmatrix}{\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{vmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w&\quad {\text{cross product}}\\\end{aligned}}

Thus, ${\frac {\Delta V_{1}}{\Delta V_{2}}}={\frac {{\begin{vmatrix}{\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}&{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}&{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&{\frac {\partial z}{\partial w}}\\\end{vmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w}{\Delta u\Delta v\Delta w}}={\begin{vmatrix}{\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}&{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}&{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&{\frac {\partial z}{\partial w}}\\\end{vmatrix}}\end{vmatrix}}$ .

Definition. (The Jacobian for three variables) The Jacobian of the transformation $T$  given by functions $x=x(u,v,w),y=y(u,v,w)$  and $z=z(u,v,w)$  whose partial derivatives exist and are continuous, is

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

The absolute value is added to prevent a negative volume.

{\begin{aligned}\iiint _{R}f(x,y,z)dV&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x_{i},y_{j},z_{k})\Delta V\\&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x(u_{i},v_{j},w_{k}),y(u_{i},v_{j},w_{k}),z(u_{i},v_{j},w_{k}))\Delta V_{2}\\{\text{Since }}&\Delta V_{2}\approx \left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\Delta u\Delta v\Delta w\\\sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x(u_{i},v_{j},w_{k}),y(u_{i},v_{j},w_{k}),z(u_{i},v_{j},w_{k}))\Delta V_{2}&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x(u_{i},v_{j},w_{k}),y(u_{i},v_{j},w_{k}),z(u_{i},v_{j},w_{k}))\ \left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\Delta u\Delta v\Delta w\\&\approx \iiint _{S}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\\\end{aligned}}

Then, we have the following theorem which is analogous to the theorem for double integrals. Again, we should aware that the above explanations are not proof of this theorem.

Theorem. (Change of variables for triple integration)Suppose $T$  is a $C^{1}$  transformation whose Jacobian is nonzero and that maps a region $S$  in the $uvw$ -space onto a region $R$  in the $xyz$ -space injectively, via the change of variables $x=x(u,v,w),y=y(u,v,w)$  and $z=z(u,v,w)$ . Suppose that $f$  is continuous on $R$ , we have

$\iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}f(x(u,v,w),y(u,v,w),z(u,v,w))\left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\,du\,dv\,dw$

Remark.
• $dx\,dy\,dz$  has the same meaning as $dV$ .
• If we change some notations, we can get

$\iiint _{S}f(u(x,y,z),v(x,y,z),w(x,y,z))\left|{\frac {\partial (u,v,w)}{\partial (x,y,z)}}\right|\,dx\,dy\,dz=\iiint _{R}f(u,v,w)\,du\,dv\,dw.$

($T$  maps a region $S$  in $xyz$ -space onto a region $R$  in $uvw$ -space in this case.) which may be a more convenient form to be used sometimes.

Example.

1 Choose correct statement(s) from the following statements.

 ${\frac {\partial (x,y,u)}{\partial (x,y,v)}}={\frac {\partial u}{\partial v}}.$ $\iiint _{D}xyz\left|{\frac {\partial (u,v,w)}{\partial (x,y,z)}}\right|\,dx\,dy\,dz$ gives the 4-dimensional volume under the graph of $f(x,y,z)=xyz$ over the region $D$ in $xyz$ -space. $\iiint _{D}(uvw)\,du\,dv\,dw$ gives the 4-dimensional volume under the graph of $f(u,v,w)=uvw$ over the region $D$ in $uvw$ -space. $\iiint _{D}\left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\,du\,dv\,dw$ gives the volume of the region $D'$ in $xyz$ -space, that is mapped from the region $D$ in $uvw$ -space by a transformation satisfying the conditions mentioned in the theorem about change of variables in a triple integral.

Now, we understand the purpose and the derivation of the Jacobian. It is time to apply this new knowledge to some examples. The first two examples consist of the change of coordinates from the Cartesian coordinate system into the polar coordinate system and the change of Cartesian to spherical coordinates.

### Change of coordinate system

Sometimes, we may change the region we are integrating over to another region in other coordinate system. This can simplify the computation of integrals, especially when the region in Cartesian coordinate system is related to circle, e.g. sphere, cone, circle, etc.

Let us start with the change of coordinates from the Cartesian coordinate system into the polar coordinate system.

Proposition. (Changing Cartesian coordinate system to polar coordinate system for double integration) Let $f(x,y)$  be a continuous function defined using Cartesian coordinates, and let $g(r,\theta )=f(r\cos \theta ,r\sin \theta )$  be the same function expressed using polar coordinates. Suppose the region $S$  in the polar coordinates is mapped injectively to the region $R$  in the Cartesian coordinates. Then,

$\iint _{R}f(x,y)\,dx\,dy=\iint _{S}g(r,\theta ){\color {green}{r}}\,dr\,d\theta .$

Proof. If we change from Cartesian coordinate system to polar coordinate system, we have the relationships

$x=r\cos \theta {\text{ and }}y=r\sin \theta .$

Thus, the Jacobian is
${\frac {\partial (x,y)}{\partial (r,\theta )}}={\begin{vmatrix}{\frac {\partial x}{\partial r}}&{\frac {\partial x}{\partial \theta }}\\{\frac {\partial y}{\partial r}}&{\frac {\partial y}{\partial \theta }}\end{vmatrix}}={\begin{vmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{vmatrix}}=r\left(\cos ^{2}\theta +\sin ^{2}\theta \right)=r.$

By the theorem about change of variables for double integration,
$\iint _{R}f(x,y)\,dx\,dy=\iint _{S}f(r\cos \theta ,r\sin \theta )\left|{\frac {\partial (x,y)}{\partial (r,\theta )}}\right|\,dr\,d\theta =\iint _{S}g(r,\theta )r\,dr\,d\theta .$

$\Box$

Proposition. (Changing Cartesian coordinate system to cylindrical coordinate system for triple integration) Let $f(x,y,z)$  be a continuous function defined using Cartesian coordinates, and let $g(r,\theta ,z)=f(r\cos \theta ,r\sin \theta ,z)$  be the same function expressed using cylindrical coordinates. Suppose the region $S$  in the cylindrical coordinates is mapped injectively to the region $R$  in the Cartesian coordinates. Then,

$\iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}g(r,\theta ,z){\color {green}{r}}\,dr\,d\theta \,dz.$

Proposition. (Changing Cartesian coordinate system to spherical coordinate system for triple integration) Let $f(x,y,z)$  be a continuous function defined using Cartesian coordinates, and let $g(\rho ,\phi ,\theta )=f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi )$  be the same function expressed using spherical coordinates. Suppose the region $S$  in the spherical coordinates is mapped injectively to the region $R$  in the Cartesian coordinates. Then,

$\iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}g(\rho ,\phi ,\theta ){\color {green}{\rho ^{2}\sin \phi }}\,d\rho \,d\phi \,d\theta .$

Proof.

Illustration of Rule of Sarrus. Red arrows correspond to the positive terms, and blue arrows correspond to the negative terms.

If we change from Cartesian coordinates to spherical coordinates, we have the relationships

$x=\rho \sin \phi \cos \theta ,\,y=\rho \sin \phi \sin \theta \;{\text{and}}\;z=\rho \cos \phi .$

Thus, the Jacobian is
{\begin{aligned}{\frac {\partial (x,y,z)}{\partial (\rho ,\phi ,\theta )}}&={\begin{vmatrix}{\frac {\partial x}{\partial \rho }}&{\frac {\partial x}{\partial \phi }}&{\frac {\partial x}{\partial \theta }}\\{\frac {\partial y}{\partial \rho }}&{\frac {\partial y}{\partial \phi }}&{\frac {\partial y}{\partial \theta }}\\{\frac {\partial z}{\partial \rho }}&{\frac {\partial z}{\partial \phi }}&{\frac {\partial z}{\partial \theta }}\\\end{vmatrix}}\\&={\begin{vmatrix}\sin \phi \cos \theta &\rho \cos \phi \cos \theta &-\rho \sin \phi \sin \theta \\\sin \phi \sin \theta &\rho \cos \phi \sin \theta &\rho \sin \phi \sin \theta \\\cos \phi &-\rho \sin \phi &0\end{vmatrix}}\\&=0{\color {purple}+\rho ^{2}\sin \phi \cos ^{2}\phi \cos ^{2}\theta }{\color {brown}+\rho ^{2}\sin ^{3}\phi \sin ^{2}\theta }{\color {purple}-(-\rho ^{2}\sin \phi \cos ^{2}\phi \sin ^{2}\theta )}{\color {brown}-(-\rho ^{2}\sin ^{3}\phi \cos ^{2}\theta )}-0\qquad {\text{by Rule of Sarrus}}\\&={\color {purple}\rho ^{2}\sin \phi \cos ^{2}\phi (\underbrace {\sin ^{2}\theta +\cos ^{2}\theta } _{1})}{\color {brown}+\rho ^{2}\underbrace {\sin ^{3}\phi } _{\sin \phi \sin ^{2}\phi }(\underbrace {\sin ^{2}\theta +\cos ^{2}\theta } _{1})}\\&=\rho ^{2}\sin \phi \left(\underbrace {{\color {purple}\cos ^{2}\phi }{\color {brown}+\sin ^{2}\phi }} _{1}\right)\\&=\rho ^{2}\sin \phi .\end{aligned}}

By the theorem about change of variables for triple integration,
$\iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi )\left|{\frac {\partial (x,y,z)}{\partial (\rho ,\phi ,\theta )}}\right|\,d\rho \,d\phi \,d\theta =\iiint _{S}g(\rho ,\phi ,\theta )\underbrace {\rho ^{2}} _{\geq 0}\overbrace {\sin \phi } ^{\geq 0\;{\text{since}}\;0\leq \phi \leq \pi }\,d\rho \,d\phi \,d\theta .$

$\Box$
Proof.

If we change from Cartesian coordinate system to cylindrical coordinate system, we have the relationships

$x=r\cos \theta ,\,y=r\sin \theta \;{\text{and}}\;{\color {green}{z=z}}.$

Thus, the Jacobian is
${\frac {\partial (x,y,{\color {green}z})}{\partial (r,\theta ,{\color {green}z})}}={\begin{vmatrix}{\frac {\partial x}{\partial r}}&{\frac {\partial x}{\partial \theta }}&{\color {green}{\frac {\partial x}{\partial z}}}\\{\frac {\partial y}{\partial r}}&{\frac {\partial y}{\partial \theta }}&{\color {green}{\frac {\partial y}{\partial z}}}\\{\color {green}{\frac {\partial z}{\partial r}}}&{\color {green}{\frac {\partial z}{\partial \theta }}}&{\color {green}{\frac {\partial z}{\partial z}}}\end{vmatrix}}={\begin{vmatrix}\cos \theta &-r\sin \theta &{\color {green}{0}}\\\sin \theta &r\cos \theta &{\color {green}{0}}\\{\color {green}0}&{\color {green}0}&{\color {green}1}\end{vmatrix}}=\underbrace {0-0+{\begin{vmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{vmatrix}}} _{\text{by cofactor expansion along 3rd row}}=r.$

By the theorem about change of variables for triple integration,
$\iiint _{R}f(x,y,{\color {green}z})\,dx\,dy\,{\color {green}dz}=\iiint _{S}f(r\cos \theta ,r\sin \theta ,{\color {green}z})\left|{\frac {\partial (x,y,{\color {green}z})}{\partial (r,\theta ,{\color {green}z})}}\right|\,dr\,d\theta \,{\color {green}dz}=\iiint _{S}g(r,\theta ,{\color {green}z})r\,dr\,d\theta \,{\color {green}dz}.$

$\Box$

Example. (Volume of the cone) Prove that the volume of a cone with radius $a$  and height $h$  is ${\frac {1}{3}}\pi a^{2}h$  by triple integration. (Hint: You may put the base of the cone on the $xy$ -plane with centre $(0,0,0)$ , point the cone to the direction of positive $z$ -axis, and use cylindrical coordinates)

Proof.

First, put the cone as instructed in the hint. Let the region bounded by the cone in Cartesian coordinate system and cylindrical coordinate system be $C$  and $C'$  respectively. Then, using cylindrical coordinates, by the proposition about triple integration using cylindrical coordinates and the proposition about volume given by triple integration, the desired volume is

$\iiint _{C}1\,dV=\iiint _{C'}r\,dr\,d\theta \,dz.$

Next, we need to find the bounds for $r,\theta$  and $z$  in the region $C'$ .

First, the bounds for $\theta$  is $0\leq \theta \leq 2\pi$  (by the definition of cylindrical coordinates).

Then, given a fixed $\theta$ , we consider the corresponding $rz$ -plane to see whether we can obtain any relationship between $r$  and $z$ . Since the region in the $rz$ -plane (it is $xz$ -plane in Cartesian coordinate system when $\theta =0$ ) over which the integral is taken is the triangle with vertices $(0,0),\,(0,h)$  and $(a,0)$ , for which the equation of the region is

${\color {green}z}\leq {\frac {-h}{a}}{\color {green}r}+h\implies {\frac {\color {green}z}{h}}\leq -{\frac {\color {green}r}{a}}+1\implies {\frac {\color {green}r}{a}}+{\frac {\color {green}z}{h}}\leq 1$

Therefore, given a fixed $\theta$
${\frac {r}{a}}\leq {\frac {r}{a}}+\underbrace {\frac {z}{h}} _{\geq 0}\leq 1\implies \underbrace {0\leq } _{\text{by definition}}r\leq a,$

(this shows that $r$  is actually independent from $\theta$ .) and given fixed $r,\theta$ ,
${\frac {z}{h}}+{\frac {r}{a}}\leq 1\implies \underbrace {0\leq z} _{\text{by cone}}\leq h\left(1-{\frac {r}{a}}\right).$

(this shows that $z$  is actually independent from $\theta$ .) Therefore, the desired volume is
{\begin{aligned}\iiint _{C'}r\,dr\,d\theta \,dz&=\int _{0}^{2\pi }\int _{0}^{a}\int _{0}^{h\left(1-{\frac {r}{a}}\right)}r\,dz\,dr\,d\theta \qquad {\text{by generalized Fubini's theorem for triple integration}}\\&=\int _{0}^{2\pi }\int _{0}^{a}rh\left(1-{\frac {r}{a}}\right)\,dr\,d\theta \\&=\int _{0}^{2\pi }\left[{\frac {hr^{2}}{2}}-{\frac {hr^{3}}{3a}}\right]_{r=0}^{r=a}\,d\theta \\&=h\int _{0}^{2\pi }\left(\underbrace {{\frac {a^{2}}{2}}-{\frac {a^{{\cancel {3}}2}}{3{\cancel {a}}}}} _{a^{2}/6}\right)\,d\theta \\&={\frac {(2\pi )a^{2}}{6}}\\&={\frac {1}{3}}\pi a^{2}h\end{aligned}}

$\Box$

Example. (Volume of the sphere) Prove that the volume of a sphere of radius $r$  is ${\frac {4}{3}}\pi r^{3}$  by triple integration. (Hint: You may put the centre of the sphere to the origin, i.e., $(0,0,0)$ .)

Proof.

First, put the centre of the sphere to the origin. Let $S$  and $S'$  be the region bounded by the sphere in Cartesian coordinate system and spherical coordinate system respectively. Using spherical coordinates, by the proposition about triple integration using spherical coordinates, the desired volume is

$\iiint _{S}1\,dV=\iiint _{S'}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta .$

Since the bounds for $\rho ,\phi ,\theta$  are $0\leq \rho \leq r,\,0\leq \phi \leq \pi$  and $0\leq \theta \leq 2\pi$  (by the definition of spherical coordinates) in the region $S'$ , the desired volume is
{\begin{aligned}\iiint _{S'}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta &=\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{r}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta \\&={\frac {1}{3}}\int _{0}^{2\pi }\int _{0}^{\pi }r^{3}\sin \phi \,d\phi \,d\theta \\&={\frac {1}{3}}r^{3}\int _{0}^{2\pi }\overbrace {-(\underbrace {\cos(\pi )} _{-1}-\underbrace {\cos 0} _{1})} ^{2}\,d\theta \\&={\frac {2}{3}}r^{3}(2\pi )\\&={\frac {4}{3}}\pi r^{3}.\end{aligned}}