# Multivariate CoV

Polar coordinates are one example of a change of variables in multivariate integration. Other important examples of this arise in triple integrals, including cylindrical and spherical coordinates. Before we do that, it might make sense to talk about change of variables in general.

## An elliptic example

What is the volume below the graph of $$f(x,y)=4-(x^2+4y^2)$$ and over the $xy$-plane, as shown below? Note that this image is very similar to the polar image we discussed last time.

### Polar coordinates don't help

Note that polar coordinates don't help here, since $$4- ((r\cos(\theta))^2 + 4(r\sin(\theta))^2)$$ doesn't simplify in any meaningful way.

### New coordinates

To describe the elliptical domain, we're going to introduce new variables $s$ and $t$ by defining their relationship with $x$ and $y$:

\begin{align} x &= 2\,s\cos(t)\\ y &= s\sin(t). \end{align}

For a fixed value of $s$, these equations parametrize an ellipse that's twice as wide as it is tall. For the set of all values of $s$ from zero to one, we trace out a solid ellipse.

### New coordinates (cont)

We might think that this new coordinate could help because the function simplifies under the transformation:

\begin{align} 4-(x^2+4y^2) &= 4-\left((2s\cos(t))^2 + 4(s\sin(t))^2\right)\\ &=4-4s^2(\cos^2(t)+\sin^2(t)) = 4-4s^2. \end{align}

Bringing us to:

$$\iint_{E} \left(4-(x^2+4y^2)\right) dA = \int_{\color{red}?}^{\color{red}?} \int_{\color{red}?}^{\color{red}?} (4-4s^2)\,{\color{red}?}\,ds\,dt.$$

### New coordinates (cont 2)

from a geometric perspective, we might say that the parametric equations $$x = 2\,s\cos(t), \: \: y = s\sin(t)$$ map the rectangle $[0,1]\times[0,2\pi]$ in the $st$-plane to the ellipse in the $xy$-plane. ### Almost there

At this point, we've got

$$\iint_{E} \left(4-(x^2+4y^2)\right) dA = \int_{0}^{2\pi} \int_0^1 (4-4s^2)\,{\color{red}?}\,ds\,dt.$$

We've just got to figure out that $\color{red}?$.

## The last piece

The last piece of the puzzle is the $\color{red}?$ in

$$\int_0^{2\pi} \int_0^1 (4-4s^2)\,{\color{red}?}\,ds\,dt.$$

It should be exactly analogous to the $\color{red}r$ in polar coordinates:

$$\int_{\alpha}^{\beta} \int_a^b F(r,\theta)\,{\color{red}r}\,dr\,d\theta.$$

### Last piece (cont)

Recall that the $r$ in the $r\,dr\,d\theta$ of polar coordinates accounts for the fact that the lines of constant $\theta$ and arcs of constant $r$ don't break a polar region doesn't break up evenly. In the context of Change of Variables, this boils down to the fact that the mapping from one coordinate system to the other distorts area.

### Last piece (cont 2)

Now our transformation, $x=2s\cos(t)$ and $y=s\sin(t)$, also distorts area, but it distorts in the $x$ direction by twice as much. So, perhaps $\displaystyle dA = 2s\,ds\,dt$?

### Last piece (cont 3)

Assuming that works, we can now finish off the integral:

\begin{align} \iint_{E} \left(4-(x^2+4y^2)\right) dA &= \int_0^{2\pi} \int_0^1 (4-4s^2)\,2s\,ds\,dt \\ &= 16\pi\int_0^1(s-s^3)\,ds = 4\pi. \end{align}

## The Jacobian

When we have a function (often called a transformation in this context) $T:\mathbb R^2\to\mathbb R^2$ there is a a standard tool called the Jacobian used to measure how it distorts area. Write $T$ componentwise:

$$T(s,t) = (x(s,t),y(s,t)), \text{ then}$$ $$JT = \left|\frac{\partial(x,y)}{\partial(s,t)}\right| = \left|\Big| \begin{matrix} \partial x/\partial s & \partial x/\partial t \\ \partial y/\partial s & \partial y/\partial t \end{matrix} \Big|\right|.$$

That's a determinant inside an absolute value.

### Jacobian example

Let's find the Jacobian for the general elliptic coordinates:

$$x = R_1\,s\,\cos(t) \: \text{ and } \: y = R_2\,s\sin(t).$$ \begin{align} \left|\frac{\partial(x,y)}{\partial(s,t)}\right| &= \left|\begin{matrix} R_1\cos(t) & -R_1\,s\,\sin(t) \\ R_2\sin(t) & R_2\,s\cos(t)\end{matrix} \right| \\ &= R_1R_2s\cos^2(t) + R_1R_2s\sin^2(t) = R_1R_2s. \end{align}

This agrees with our prior example where $R_1=2$ and $R_2=1$.

## Integrating over a parallelogram

Let $P$ denote the parallelogram bound by the lines

$$\begin{array}{cccc} 3x-y=0 & 3x-y=8 & 3y-x=0 & 3y-x=8 \end{array}$$ ### Parallelogram (cont)

We wish to set up a single, iterated integral representing

$$\iint_P xy\,dA.$$

To do so, introduce $s$ and $t$ defined by

$$s = 3x-y \: \text{ and } \: t = 3y-x.$$

For then, $0<s<8$ and $0<t<8$. Solving, we get

$$x = (3 s + t)/8 \: \text{ and } \: y = (s + 3 t)/8.$$

### Parallelogram (cont 2)

The Jacobian is

$$\left|\frac{\partial(x,y)}{\partial(s,t)}\right| = \left|\begin{matrix} 3/8 & 1/8 \\ 1/8 & 3/8\end{matrix} \right| = \frac{1}{8}.$$

Thus, the integral is

\begin{align} \iint_{p} xy \, dA &= \frac{1}{8}\int_0^{8} \int_0^8 \frac{3s+t}{8}\frac{3t+s}{8}\,ds\,dt. \end{align}

## Last virtual example!

As a final example, let's find the area of the region $R$ bound between the curves $$y=2x, \: y=\frac{1}{2} x, \: y=\frac{1}{x}, \text{ and } y=\frac{2}{x}.$$ ### Last (cont)

We wish to set up a single, iterated integral representing

$$\iint_R 1 \, dA.$$

To do so, introduce $s$ and $t$ defined by

$$s = y/x \: \text{ and } \: t = xy.$$

For then, $1/2<s<2$ and $1<t<2$.

### Last (cont 2)

Since, $s=y/x$, we have $y=sx$. We can plug that into the other equation ($t=xy$) to get $t=sx^2$. Thus, $$x=\sqrt{t/s}.$$ Once we know that, we can find that $$y = s\sqrt{t/s} = \sqrt{st}.$$

### Last (cont 3)

Since $$x = \sqrt{t/s} = s^{-1/2}t^{1/2} \: \text{ and } \: y = \sqrt{s t} = s^{1/2}t^{1/2},$$

the Jacobian is

$$\left|\frac{\partial(x,y)}{\partial(s,t)}\right| = \left|\begin{matrix} -\frac{1}{2}s^{-3/2}t^{1/2} & \frac{1}{2}s^{-1/2}t^{-1/2} \\ \frac{1}{2}s^{-1/2}t^{1/2} & \frac{1}{2}s^{1/2}t^{-1/2} \end{matrix} \right| = -\frac{1}{4s}-\frac{1}{4s} = -\frac{1}{2s}.$$

Don't forget to take the absolute value to get $\frac{1}{2s}$.

Putting this all together, the area is $$\iint_R 1 \, dA = \int_1^2 \int_{1/2}^2 1\,\frac{1}{2s} \, ds \, dt = \ln(2).$$