## The Jacobian

The following image is Image 16.57 from your text, but I’ve modified it slightly. The idea is that we’re starting off in the situation on the left, with a function given in terms of $x$ and $y$ and a region $R$ awkwardly written (in this case a parallelogram) in the Cartesian plane. When we perform our change of variables, we are now integrating over a region $T$ in the variables $s$ and $t$. The key thing to notice is that $T$ is a rectangle, and we like integrating over rectangles.

For instance, this is the situation when we take a double integral in the Cartesian plane with $x$ and $y$ and convert it to a double integral in polar coordinates. Notice that an integral such as
$\int_{0}^{2\pi}\int_0^1 r\,dr\,d\theta$
calculates the area of a circle, by integrating over the rectangle where $0\le r\le 1$ and $0\le\theta\le 2\pi$. The extra $r$ in the integrand tells us that area increases linearly as $r$ increases from left to right inside this rectangle. So, looking at $T$ in the figure above, a small rectangle at the left side of $T$ is worth less in terms of area than the same sized rectangle at the right side of $T$.

In a general situation, we’d like to calculate our integral in the variables $s$ and $t$ given by the region $T$. But, we need to know how area in $x$ and $y$ relates to area in $s$ and $t$. In polar coordinates, the area in $R$ is given by taking the area in $T$ and multiplying it by $r$ (and so the area becomes more valuable as we move from left to right in $T$). In the general situation, we must calculate this translation factor, which is known as a Jacobian.

We view $x$ as a function $x=x(s,t)$ of $s$ and $t$. For instance, if $s=r$ and $t=\theta$, then $x=r\cos\theta$. Likewise, we view $y$ as a function $y(s,t)$. Given the diagram above, we have

\begin{align*} \vec{a} &= \left(x(s+\Delta s,t)-x(s,t)\right)\vec{i}+\left(y(s+\Delta s,t)-y(s,t)\right)\vec{j}\\ \vec{b} &= \left(x(s,t+\Delta t)-x(s,t)\right)\vec{i}+\left(y(s,t+\Delta t)-y(s,t)\right)\vec{j}.\\ \end{align*}

Notice that each component of $\vec{a}$ and $\vec{b}$ almost looks like a derivative. In fact, if we placed a $\Delta s$ or $\Delta t$ in the denominator, we’d have a difference quotient. So, we’ll multiply $\vec{a}$ by $\Delta s/\Delta s$ and multiply $\vec{b}$ by $\Delta t/\Delta t$, giving

$\vec{a}=\frac{\partial x}{\partial s}\Delta s\vec{i}+\frac{\partial y}{\partial s}\Delta s\vec{j} \qquad\text{and}\qquad \vec{b}=\frac{\partial x}{\partial t}\Delta t\vec{i}+\frac{\partial y}{\partial t}\Delta t\vec{j}.$

The whole reason why we’re doing this is to make the calculation of $\vec{a}\times\vec{b}$ easier to compute, since the cross product represents the area of the parallelogram between the two vectors. (We know the area in $T$ with respect to $s$ and $t$, and we want to know the corresponding area in $R$ with respect to $x$ and $y$.) We now compute the cross product. Actually, there are some technicalities here.

1. We only compute the magnitude of the cross product, which represents the area of the parallelogram. Recall that the cross product (which is only defined for 3-dimensional vectors) is a 3-dimensional vector. We’re only interested in the magnitude of the cross product, not its direction.

2. We could use the geometric interpretation $|\vec{a}\times\vec{b}|=|\vec{a}|\cdot|\vec{b}|\sin\theta$, if we knew the angle $\theta$ between $\vec{a}$ and $\vec{b}$. Usually, we don’t know $\theta$. So, we use the algebraic interpretation of the cross product:

$\vec{a}\times\vec{b}=\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{array} \right|$

where we’re taking the determinant of the matrix and $a_3=b_3=0$ (since $\vec{a}$ and $\vec{b}$) are vectors in the $xy$-plane with no $z$-component).

OK, now we calculate the magnitude of the cross product.

\begin{align*} |\vec{a}\times\vec{b}| &= \left|\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k}\\ & & \\ \displaystyle\frac{\partial x}{\partial s}\Delta s &\displaystyle \frac{\partial y}{\partial s}\Delta s & 0\\ & & \\ \displaystyle\frac{\partial x}{\partial t}\Delta t & \displaystyle\frac{\partial y}{\partial t}\Delta t & 0 \end{array} \right|\right|\\ &= \left|\frac{\partial x}{\partial s}\Delta s\frac{\partial y}{\partial t}\Delta t-\frac{\partial x}{\partial t}\Delta t\frac{\partial y}{\partial s}\Delta s\right|\\ &=\Delta s\Delta t\left|\frac{\partial x}{\partial s}\frac{\partial y}{\partial t}-\frac{\partial x}{\partial t}\frac{\partial y}{\partial s}\right| \end{align*}

The idea here is that $\Delta s\Delta t$ represents area in a rectangle in variables $s$ and $t$, while the partials computation that follows represents the translation factor for area in $x$ and $y$. Unfortunately, this formula is hard to memorize. With that in mind, we note that

$|\vec{a}\times\vec{b}| = \left|\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k}\\ & & \\ \displaystyle\frac{\partial x}{\partial s}\Delta s &\displaystyle \frac{\partial y}{\partial s}\Delta s & 0\\ & & \\ \displaystyle\frac{\partial x}{\partial t}\Delta t & \displaystyle\frac{\partial y}{\partial t}\Delta t & 0 \end{array}\right|\right| = \left|\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k}\\ & & \\ \displaystyle\frac{\partial x}{\partial s}\Delta s & \displaystyle\frac{\partial x}{\partial t}\Delta t & 0\\ & & \\ \displaystyle \frac{\partial y}{\partial s}\Delta s & \displaystyle\frac{\partial y}{\partial t}\Delta t & 0 \end{array}\right|\right| = \left|\left| \begin{array}{cc} \displaystyle\frac{\partial x}{\partial s} & \displaystyle\frac{\partial x}{\partial t}\\ & \\ \displaystyle \frac{\partial y}{\partial s} & \displaystyle\frac{\partial y}{\partial t} \end{array} \right|\right|\Delta s\Delta t.$

Thus, we define the Jacobian using this slightly easier to remember formula:

$\frac{\partial(x,y)}{\partial(s,t)}=\left|\left|\begin{array}{cc} \displaystyle\frac{\partial x}{\partial s} & \displaystyle\frac{\partial x}{\partial t}\\ & \\ \displaystyle \frac{\partial y}{\partial s} & \displaystyle\frac{\partial y}{\partial t} \end{array} \right|\right|\,ds\,dt$

## Algorithm for 2-Dimensional Change of Variables

To change the variables of a double integral $\displaystyle\int\int_R f(x,y)\,dx\,dy$ from the $xy$-plane to the $st$-plane:

1. Write the function $f(x,y)$ as a function in $s$ and $t$.

2. Write the bounds of integration in terms of $s$ and $t$.

3. Calculate the Jacobian and use it to replace $dx\,dy$ as

$dx\,dy = \frac{\partial(x,y)}{\partial(s,t)}= \left|\left|\begin{array}{cc} \displaystyle\frac{\partial x}{\partial s} & \displaystyle\frac{\partial x}{\partial t}\\ & \\ \displaystyle \frac{\partial y}{\partial s} & \displaystyle\frac{\partial y}{\partial t} \end{array} \right|\right|\,ds\,dt$

## Polar Coordinates

Problem: Calculate the area of a circle of radius 1 centered at the origin.

Solution: We wish to calculate

$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}1\,dy\,dx.$

Step (1), changing the function $f(x,y)=1$ to be in terms of $r$ and $\theta$ is trivial. We still have 1 instead the integral. Step (2) is where we change the limits of integration to $0\le \theta\le 2\pi$ and $0\le r\le 1$. Step (3) is where we calculate

$dx\,dy = \frac{\partial(x,y)}{\partial(r,\theta)}= \left|\left|\begin{array}{cc} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{array} \right|\right|\,dr\,d\theta= (r\cos^2\theta +r\sin^2\theta)\,dr\,d\theta= r\,dr\,d\theta$

Putting this together, we have

$\int_0^{2\pi}\int_0^1 1r\,dr\,d\theta=\pi.$

## Area of an Ellipse

This is an example taken from our text, with some more details added. Find the area of the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1.$

Solution: I would approach this a different way than your text does. First, I’m only interested in finding the area in one quadrant, since the area of the entire ellipse is four times that. Solving for $y$ and looking at the first quadrant gives us

$4\int_0^a\int_0^{\sqrt{b^2-\frac{b^2x^2}{a^2}}} 1\,dy\,dx.$

That integral doesn’t look fun. Let’s use a change of variables to squash the ellipse into a circle. We’ll use the variables $s$ and $t$ and write our original variables $x$ and $y$ in terms

$x=as\qquad\text{and}\qquad y=bt.$

Then, we are looking at finding the area of

$s^2+t^2\le 1$

in the $st$-plane. The function we’re integrating is 1, and so step (1) of the algorithm is already complete for us. Step (2) involves changing the limits of integration. Again, we’re only going to consider one quadrant, and multiply the result by 2. The Jacobian turns out to be $ab\,ds\,dt$. All of this together gives us

$4\int_0^1\int_0^{\sqrt{1-t^2}}ab\,ds\,dt=4ab\int_0^1\int_0^{\sqrt{1-t^2}}1\,ds\,dt.$

Converting this again to polar coordinates (and going through all 3 steps) gives us

$4ab\int_0^1\int_0^{\sqrt{1-t^2}}1\,ds\,dt=4ab\int_0^{\pi/2}\int_0^1r\,dr\,d\theta.$

Calculating this yields

$4ab\int_0^{\pi/2}\int_0^1r\,dr\,d\theta=4ab\int_0^{\pi/2}\,d\theta\int_0^1r\,dr=4ab\frac{\pi}{2}\frac{1}{2}=\pi ab.$