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

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

it would be helpful if we can substitute ${\displaystyle x+y}$ as ${\displaystyle u}$ and ${\displaystyle x-y}$ as ${\displaystyle v}$ because ${\displaystyle e^{\frac {u}{v}}}$ 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.

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

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

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

However, the area of ${\displaystyle R}$ is too complicated to be written out in terms of ${\displaystyle x,y}$. So, we want to change the variables so that the area of ${\displaystyle R}$ 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 ${\displaystyle u,v}$ that have connections with variables ${\displaystyle x,y}$ that satisfy:

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

The original integral can be rewritten into:

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

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

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

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 ${\displaystyle x,y,z}$ into ${\displaystyle r,\theta ,z}$ with relations:

${\displaystyle x=r\cos \theta \quad y=r\sin \theta \quad z=z}$

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

${\displaystyle \iiint _{R}f(x,y,z)dV=\iiint _{S}f(r\cos \theta ,r\sin \theta ,z)dz\ r\ dr\ d\theta }$, 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 ${\displaystyle x,y,z}$ and the transformed variables ${\displaystyle \rho ,\theta ,\phi }$ have the relations:

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

These relations can give us that

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

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 ${\displaystyle x=r\cos \theta ,y=r\sin \theta {\text{ and }}z=z}$, we will talk about general changes. Let's start from two variables.

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

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

In this transformation, ${\displaystyle x,y}$ are related to ${\displaystyle u,v}$ by the equations

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

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

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

$${\displaystyle T(S)=R}$$

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

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

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

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

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

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

${\displaystyle u=0\quad u=1\quad v=0\quad v=1}$

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

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

We can use the same method to calculate ${\displaystyle S}$ from ${\displaystyle R}$.

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 ${\displaystyle \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|}$ from ${\displaystyle \iint _{S}f(x(u,v),y(u,v))\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\ du\ dv}$ here. To actually start explaining that, we should review some basic concepts.

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

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

For example,

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

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

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

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

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

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

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

The appearance of the term ${\displaystyle {\frac {dx}{du}}}$ not only is a mathematical product of deduction, but also serves a intuitive purpose. When we change our function from ${\displaystyle f(x)}$ to ${\displaystyle f(x(u))}$, 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 ${\displaystyle {\frac {du}{dx}}}$. To counter this change, ${\displaystyle {\frac {dx}{du}}}$ is deduced to compromise (recall that ${\displaystyle {\frac {dx}{du}}\left({\frac {du}{dx}}\right)=1}$). 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 ${\displaystyle x,y}$ to ${\displaystyle u,v}$, 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:

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

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 ${\displaystyle f(x,y)}$ to ${\displaystyle f(x(u),y(u))}$, we "stretched" or "condensed" the area of our region, by a factor of ${\displaystyle {\frac {dA_{2}}{dA_{1}}}}$; therefore, we need to counter the change with a factor of ${\displaystyle {\frac {dA_{1}}{dA_{2}}}}$. The Jacobian matrix for two variables is basically an expression for calculating ${\displaystyle {\frac {dA_{1}}{dA_{2}}}}$ in terms of ${\displaystyle u,v}$, 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 ${\displaystyle u,v}$, but not ${\displaystyle x,y}$ (we need to express ${\displaystyle x}$ and ${\displaystyle y}$ in terms of ${\displaystyle u,v}$).

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 ${\displaystyle {\frac {dA_{1}}{dA_{2}}}}$, with ${\displaystyle dA_{1}}$ being the infinitesimally small area in the region ${\displaystyle R}$ in the ${\displaystyle xy}$-plane and ${\displaystyle dA_{2}}$ being the infinitesimally small area in the region ${\displaystyle S}$ in the ${\displaystyle uv}$-plane. Since we are changing our variables from ${\displaystyle x,y}$ to ${\displaystyle u,v}$, we should describe ${\displaystyle dA_{1}}$ and ${\displaystyle dA_{2}}$ in terms of ${\displaystyle u,v}$ over a region in ${\displaystyle uv}$-plane.

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

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

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

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

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

The shape of the region ${\displaystyle R_{0}}$ after transformation ${\displaystyle R_{0}=T(S_{0})}$ 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 ${\displaystyle \mathbf {a} }$ and ${\displaystyle \mathbf {b} }$ can be calculated by taking the magnitude of the cross product of the two vectors.

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

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

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

It seems very similar to the definition of partial derivatives:

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

As a result, we can approximate that:

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

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

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

We can calculate ${\displaystyle \Delta A_{1}=||\mathbf {a} \times \mathbf {b} ||}$ (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.

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

Then, we can substitute our newly deduced terms.

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

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

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

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

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.

Example.

1

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

	${\displaystyle {\frac {1}{10}}}$
	${\displaystyle {\begin{vmatrix}0&{\frac {1}{2}}\\{\frac {1}{5}}&0\end{vmatrix}}}$
	${\displaystyle {\begin{vmatrix}2&0\\0&5\end{vmatrix}}}$
	${\displaystyle {\begin{vmatrix}0&2\\5&0\end{vmatrix}}}$
	${\displaystyle 10}$

2

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

	${\displaystyle 3\int _{1}^{2}\int _{0}^{1}uv\,du\,dv}$
	${\displaystyle -3\int _{0}^{1}\int _{1}^{2}uv\,du\,dv}$
	${\displaystyle -{\frac {1}{3}}\int _{0}^{1}\int _{1}^{2}uv\,du\,dv}$
	${\displaystyle {\frac {1}{3}}\int _{1}^{2}\int _{0}^{1}uv\,du\,dv}$
	${\displaystyle {\frac {1}{3}}\int _{0}^{1}\int _{1}^{2}uv\,du\,dv}$

3

Choose correct statement(s) from the following statements.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle {\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}}}$