Switching Coordinates: The Generalized Jacobian

Section11.4Switching Coordinates: The Generalized Jacobian

Objectives

In this section you will learn how to...

switch coordinate systems for integration ( $u$ -substitution for higher dimensions)

To find the Jacobian of the transformation $\vec T(u,v)=(x(u,v), y(u,v))\text{,}$ we first find the derivative of $\vec T\text{.}$ This is a square matrix, so it has a determinant, which should give us information about area. As the determinant may be positive or negative, we then take the absolute value to obtain the Jacobian. Formally:

Definition11.4.1

Suppose $\vec T(u,v)=(x(u,v),y(u,v))$ is a differentiable coordinate transformation. The Jacobian of the transformation $\vec T$ is the absolute value of the determinant of the derivative. Notationally we write

$\begin{equation*} J(u,v) = \frac{\partial (x,y)}{\partial (u,v)} = |\det(D\vec T(u,v))|. \end{equation*}$

For a tongue twister, say “the absolute value of the determinant of the derivative” ten times really fast.

Exercise11.4.1

Consider the transformation $u=x+2y$ and $v=2x-y\text{.}$

(a)

Solve for $x$ and $y$ in terms of $u$ and $v\text{.}$

(b)

Next compute the Jacobian $\frac{\partial (x,y)}{\partial (u,v)}\text{.}$

(c)

We were give $u$ and $v$ in terms of $x$ and $y\text{,}$ so we could have directly computed $\frac{\partial (u,v)}{\partial (x,y)}\text{.}$ Do so now.

(d)

Make a conjecture about the relationship between $\frac{\partial (x,y)}{\partial (u,v)}$ and $\frac{\partial (u,v)}{\partial (x,y)}\text{.}$

Theorem11.4.2

Suppose that $f$ is integrable over a region $R_{xy}$ in the $xy$ plane. Suppose that $\vec T(u,v)=(x(u,v),y(u,v))$ is a coordinate transformation that has the Jacobian $\ds \frac{\partial (x,y)}{\partial (u,v)}\text{.}$ Suppose the region $R_{uv}$ in the $uv$ -plane corresponds to the region $R_{xy}$ in the $xy$ -plane. Provided the Jacobian is nonzero except possibly on regions with zero area, we can then write

$\begin{equation*} \iint_{R_{xy}} f(x,y) dxdy = \iint_{R_{uv}} f(x(u,v),y(u,v)) \frac{\partial (x,y)}{\partial (u,v)} dudv. \end{equation*}$

We can remember this in differential form as

$\begin{equation*} dxdy = \frac{\partial (x,y)}{\partial (u,v)} dudv. \end{equation*}$

Subsection11.4.1Jac-o-latern Practice

Let's use this to rapidly find the area inside of an ellipse.

Exercise11.4.2

Consider the region $R$ inside the ellipse $\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1\text{.}$ We'll consider the change of coordinates given by $u=(x/a)$ and $v=(y/b)\text{.}$

(a)

Draw the region $R$ in the $xy$ -plane.

(b)

After substituting $u=x/a$ and $v=y/b\text{,}$ draw the region $R_{uv}$ in the $uv$ -plane. You should have a circle.

(c)

What is the area inside this circle in the $uv$ -plane?

(d)

Solve for $x$ and $y\text{,}$ and then compute the Jacobian $\dfrac{\partial (x,y)}{\partial (u,v)}\text{.}$

(e)

Show how to get the same result from directly computing $\dfrac{\partial (u,v)}{\partial (x,y)}\text{.}$

(f)

We know the area in the $xy$ -plane of the ellipse is $\iint_{R_{xy}} dxdy\text{.}$ Use the previous theorem to switch to an integral over the region $R_{uv}\text{.}$ Then evaluate this integral by using facts about area to prove that the area in the $xy$ plane is $\pi a b\text{.}$

Hint

you don't actually have to set up any bounds, rather just reduce this to an area integral over the region $R_{uv}\text{.}$

Exercise11.4.3

Let $R$ be the region in the plane bounded by the curves:

$\begin{align*} x+2y\amp =1\\ x+2y\amp =4\\ 2x-y\amp =0\\ 2x-y\amp =8 \end{align*}$

We want to compute the integral $\iint_R xdxdy$ using a change of coordinates. [Challenge: Try doing this without looking at the steps in the next paragraph]

(a)

Draw the region $R$ in the $xy$ -plane.

(b)

Use the change of coordinates $u=x+2y$ and $v=2x-y$ to evaluate this integral.

(c)

What are the bounds for $u$ and $v\text{?}$ You'll want to solve for $x$ and $y$ in terms of $u$ and $v\text{,}$ and then you'll need a Jacobian.

(d)

Make sure you provide a sketch of the region $R_{uv}$ in the $uv$ -plane (it should be a rectangle).

Exercise11.4.4

Use the transformation $u=3x+2y$ and $v=x+4y$ to evaluate the integral

$\begin{equation*} \iint_R (3x^2+14xy+8y^2)dxdy =\iint_R (3x+2y)(x+4y)dxdy \end{equation*}$

for the region $R$ that is bounded by the lines $y=-(3/2)x+1\text{,}$ $y=-(3/2)x+3\text{,}$ $y=-(1/4)x\text{,}$ and $y=(-1/4)x+1\text{.}$

Subsection11.4.2Computational Practice

These are provided to help you achieve better skills in basic computational answers.

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