# Double integrals



My notes of the excellent lectures 16, 17 and 18 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Recall: the integral of function of one variable $$f(x)$$ corresponds to the area below the graph of $$f$$ over $$[a,b]$$.

$$\int_a^b f(x)\,dx \nonumber$$

The input domain of $$f(x)$$ is $$x$$, therefore the region of integration $$R$$ is on a line along the $$x$$-axis. Here $$x=a$$ is the lower bound, and $$x=b$$ is the upper bound. Single variable function in $$xy$$-plane

## Definition

For a function of two variables $$f(x,y)$$, the region of integration $$R$$ is bounded by a curve on the $$xy$$-plane. Using a double integral, you can find the volume between the region and a function $$z=f(x,y)$$. Volume under $$z=f(x,y)$$ over region $$R$$

To compute the volume, start with cutting the area of $$R$$ in small pieces $$\Delta A=\Delta y\Delta x$$ $$xyz$$-space with region $$R$$ and area $$\Delta A$$ $$xy$$-plane with region $$R$$ and area $$\Delta A$$

Consider all the pieces, and take the limit $$\Delta A_i\to 0$$. $$\lim_{\Delta A_i\to 0}\sum_i f(x_i,y_i)\,\Delta A_i \nonumber$$

Let $$dA=dy\,dx$$ be a tiny piece of area in region $$R$$. This gives the definition of the double integral of $$f(x,y)$$ over region $$R$$. $$\shaded{ \iint_R f(x,y)\,dA } \nonumber$$

Double integrals are evaluated as two embedded integrals, starting with the inner integral $$\int_{x_{min}}^{x_{max}} \underbrace{ \int_{y_{min}(x)}^{y_{max}(x)} f(x,y)\,dy }_{\text{function of only }y} \,dx \nonumber$$ The bound functions encode the shape of region $$R$$.

The bounds of the inner integral might be functions of the outer variables.

## In Cartesian coordinates

To compute $$\iint_R f(x,y)\,dA$$, we take slices that scan the volume from the back to the front. Slice for a given $$x_i$$ in $$xyz$$-space Slice for a given $$x$$ in $$xy$$-plane

For the outer integral, let $$S(x_i)$$ be the area of a slice $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum\ yz$$-plane (the area of the thin purple vertical wall in the picture on the left). Then, the volume of each slice is $$S(x_i)\,\Delta x$$. The total volume follows as \begin{align} \rm{volume} &= \lim_{\Delta x\to 0}\sum_i S(x)\,\Delta x \nonumber \\ &=\int_{x_{min}}^{x_{max}} \underline{S(x)}\,dx \label{eq:doublecomp1} \end{align}

For the inner integral, $$x$$ is constant and $$y$$ is the variable of integration. For the range of $$y$$, we go from the far left to the far right on the given slice, as shown in the picture on the right $$S(x) = \int_{y_{min}(x)}^{y_{max}(x)} f(x,y)\,dy \label{eq:doublecomp2}$$ Note that these inner bounds depend on $$x$$.

Substituting equation $$\eqref{eq:doublecomp2}$$ in $$\eqref{eq:doublecomp1}$$ give the iterated integral $$\shaded{ \iint_R f(x,y)\,dA = \int_{x_{min}}^{x_{max}} \left[ \int_{y_{min}(x)}^{y_{max}(x)} f(x,y)\,dy \right] dx } \nonumber$$

### Examples

#### One

Integrate $$z=1-x^2-y^2$$ over the region \left\{\begin{align*} 0\leq &x\leq 1 \\ 0\leq &y\leq 1 \end{align*}\right. \nonumber

Plot Volume under $$z=f(x,y)$$ over region $$R$$

The bounds are trivial \begin{align*} \iint_R z(x,y)\,dA &= \int_0^1\underline{\int_0^1 1-x^2-y^2\,dy}\,dx \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_0^1 1-x^2-y^2\,dy &= \left[ y-x^2y-\frac{y^3}{3} \right]_{y=0}^1 \\ &= (1-x^2-\frac{1}{3}) – 0 \\ &= \underline{\frac{2}{3}-x^2} \end{align*} \nonumber

Substituted back in the outer integral \begin{align*} \iint_R z(x,y)\,dA &=\int_0^1 \underline{\frac{2}{3}-x^2}\,dx \\ &=\left[\frac{2}{3}x-\frac{x^3}{3}\right]_{x=0}^1 = \frac{1}{3} \end{align*} \nonumber

#### Two

Integrate $$z=1-x^2-y^2$$ over the quarter unit disk region \left\{\begin{align*} x^2 + y^2 &\leq 1 \\ x &\geq 0 \\ y &\geq 0 \end{align*}\right. \nonumber

Plot $$z=f(x,y)$$ and region $$R$$ in $$xyz$$-space Region $$R$$ in $$xy$$-plane

Find the bounds of integration

1. For $$\int dy$$, the inner integral, express the bounds of $$y$$ as a function of $$x$$. The lower bond is $$0$$. The upper bounds are on a quarter circle with $$x^2+y^2 = 1 \Rightarrow y=\sqrt{1-x^2}$$.
2. For $$\int dx$$, the outer integral, the range for $$x$$ is $$0$$ to $$1$$.

Fill in the bounds of the integrals \begin{align*} \iint_R z(x,y)\,dA &= \int_0^1\underline{\int_0^{\sqrt{1-x^2}} 1-x^2-y^2\,dy}\,dx \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_0^{\sqrt{1-x^2}} 1-x^2-y^2\,dy &= \left[ y-x^2y-\frac{y^3}{3} \right]_{y=0}^{\sqrt{1-x^2}} \\ &= \left(\sqrt{1-x^2}-x^2\sqrt{1-x^2}-\frac{1}{3}(1-x^2)^{3/2}\right) – 0 \\ &= (1-x^2)(1-x^2)^{1/2}-\frac{1}{3}(1-x^2)^{3/2} \\ &= \underline{\frac{2}{3}\left(1-x^2 \right)^{3/2}} \end{align*} \nonumber

Substitute back in the outer integral \begin{align*} \iint_R z(x,y)\,dA &= \int_0^1\underline{\frac{2}{3}\left(1-x^2 \right)^{3/2}}\,dx \end{align*} \nonumber

For computing the outer integral, substitute $$x=\sin\theta$$ and using the double angle formula $$cos^2\theta=\frac{1}{2}(1+\cos2\theta)$$ twice. This will eventually lead to the answer $$\frac{\pi}{8}$$.

As we will see later, using polar coordinates will be much easier!

## Changing the order of integration

We change the order of integration, when it makes it easier to compute the double integral.

### Examples

#### One

When the bounds are numbers, they form a rectangle and we can simply switch the order of integration $$\int_0^1\int_0^2 dx\,dy = \int_0^2\int_0^1 dy\,dx \nonumber$$

#### Two

The written way can’t be computed. Change the order of integration. $$\int_0^1\int_x^{\sqrt{x}} \frac{e^y}{y}dy\,dx \nonumber$$

Plot the region based on the existing bounds. $$y=f(x)$$ $$x=f(y)$$

For the new inner integral, $$y$$ is constant and $$x$$ is the variable of integration. The old upper bound $$y=\sqrt{x} \Rightarrow x=y^2$$, and lower bound $$y=x \Rightarrow x=y$$ \begin{align*} \int_0^1\int_x^{\sqrt{x}} \frac{e^y}{y}dy\,dx &= \int_0^1 \underline{\int_{y^2}^y \frac{e^y}{y}dx}\,dy \\ \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_{y^2}^y \frac{e^y}{y}dx &= \left[x\frac{e^y}{y}\right]_{x=y^2}^y \\ &=e^y – e^y y \end{align*} \nonumber

Find the antiderivative for $$e^y – e^y y$$ (or use integrating by parts) \begin{align*} \left(y\,e^y\right)’ &= 1.e^y+y.(e^y)’=e^y+y\,e^y \\ \Rightarrow \left(-y\,e^y\right)’ &= -e^y-y\,e^y \\ \Rightarrow \left(-y\,e^y+2\,e^y\right)’ &= -e^y-y\,e^y + 2e^y \\ &= e^y-y\,e^y \end{align*} \nonumber

The outer integral evaluates to \begin{align*} \int_0^1\int_x^{\sqrt{x}} \frac{e^y}{y}dy\,dx &= \int_0^1 (e^y – e^y y)\,dx \\ &=\Big[ -y\,e^y + 2\,e^y \Big]_{y=0}^1 \\ &= (-1.e^1+2e^1)-(0+2.e^0) \\ &= e -2 \end{align*} \nonumber

#### Three

Exchange the order of integration to $$dx\,dy$$ for $$\int_0^1\int_x^{2x}f\,dy\,dx \nonumber$$

Plot the region based on the existing bounds.

These not simply connected regions results in two terms: $$0\lt y\lt 1$$ and $$1\lt y\lt 2$$. Each with different bounds for $$x$$ $$\int_0^1\int_x^{2x}f\,dy\,dx = \int_{0}^{1}\int_{y/2}^{y} f\,dx\,dy + \int_{1}^{2}\int_{y/2}^{1} f\,dx\,dy \nonumber$$

## In polar coordinates

In general, you switch to polar coordinates because the region is easier to setup, or the integrand becomes simpler. Polar-coordinates vs. $$xy$$-coordinates

Polar coordinates express point $$(x,y)$$ in the plane, using $$r$$ for the distance from the origin $$r$$, and $$\theta$$ as the counterclockwise angle with the positive $$x$$-axis. \shaded{ \begin{align*} x &= r\cos\theta \\ y &= r\sin\theta \end{align*} } \nonumber

### Area element

The area element $$\Delta A$$ is almost rectangular as shown below Area $$\Delta A$$

One side is $$\Delta r$$ and the other side is $$r\,\Delta\theta$$. For the limit where $$\Delta\theta,r\to 0$$, the area element becomes $$\shaded{ dA=r\,dr\,d\theta } \nonumber$$

The double integral in polar coordinates $$\shaded{ \int_{\theta_{min}}^{\theta_{max}} \int_{r_{min}}^{r_{max}} f(r,\theta)\,r\,dr\,d\theta } \nonumber$$

### Examples

#### One

Redo the earlier problem using polar coordinates: Integrate $$z=1-x^2-y^2$$ over the quarter unit disk region \left\{\begin{align*} x^2 + y^2 &\leq 1 \\ x &\geq 0 \\ y &\geq 0 \end{align*}\right. \nonumber

Plot of the region Region $$R$$ in polar-coordinates

Set the bounds for the integrals

1. For $$\int dr$$, the inner integral: fix the value of $$\theta$$, and let $$r$$ vary. For the bounds, ask yourself for what values of $$r$$ will I be inside my region. In this case, that is $$0\lt r\lt 1$$. We let $$\theta$$ vary.
2. For $$\int d\theta$$, the outer integral: ask yourself what values of $$\theta$$ will I be inside my region.

Fill in the bounds of the double integral $$\int_0^{\pi/2}\int_0^1 f(r,\theta)\,r\,dr\,d\theta \nonumber$$

Instead of just replacing $$x=r\,\cos\theta$$ and $$y=r\,\sin\theta$$, we can express the function $$f(x,y)$$ in polar coordinates using $$r^2=x^2+y^2$$ \begin{align*} f(x,y) &= 1-x^2-y^2 \\ &= 1-(x^2+y^2) \\ \Leftrightarrow f(r,\theta) &= 1-r^2 \end{align*} \nonumber

Evaluate the double integral \begin{align*} \text{volume} &= \int_0^{\pi/2}\underline{\int_0^1 (1-r^2)\,r\,dr}\,d\theta \\ &= \int_0^{\pi/2}\left[ \frac{r^2}{2}-\frac{r^4}{4} \right]_{r=0}^1 \,d\theta \\ &= \int_0^{\pi/2} \frac{1}{4} \,d\theta = \frac{1}{4}\frac{\pi}{2}=\frac{\pi}{8} \end{align*} \nonumber

## Applications

### Find the area

Find the area of region $$R$$. $$\shaded{ \text{Area}(R)=\iint_R 1\,dA } \nonumber$$

Or, the mass of a (flat) object with density $$\delta$$ = mass per unit area. \shaded{ \begin{align*} \Delta m &= \delta .\Delta A \\ \Rightarrow \text{Mass}(R) &= \iint_R\delta(x,y)\,dA \end{align*} } \nonumber

### Find the average value

Average value of $$f$$ in $$R$$. $$\shaded{ \bar f = \frac{1}{\text{Area}(R)}\iint_R f(x,y)\,dA } \nonumber$$

Or, the weighted average value of $$f$$ in $$R$$ with density $$\delta$$ $$\shaded{ \frac{1}{\text{Mass}(R)}\iint_R f(x,y)\,\underbrace{\delta(x,y)\,dA}_{\text{mass element}} } \nonumber$$

Or, the center of mass $$(\bar x,\bar y)$$ of a (planar) object with density $$\delta$$. The weighted averages on $$x$$ and $$y$$ \shaded{ \left\{ \begin{align*} \bar x &= \iint_R x\,\delta(x,y)\,dA \\ \bar y &= \iint_R y\,\delta(x,y)\,dA \end{align*} \right. } \nonumber

### Find the moment of inertia

Recall from physics:

The kinetic energy of a point mass equals $$\frac{1}{2}mv^2$$

Mass is how hard it is to impart a translation movement. (to make it move)

Similarly, the moment of inertia about an axis is how hard it is to rotate about that axis (to make it spin).

Let $$\omega$$ be the rate of change of angle $$\theta$$, $$\omega=\frac{d\theta}{dt}$$.

At unit time, a mass $$m$$ rotating by $$\omega$$, goes a distance of $$r\omega$$, so the speed is $$v=r\omega$$. The kinetic energy follows as $$\shaded{ \tfrac{1}{2}m\,v^2=\tfrac{1}{2}\underline{mr^2}\omega^2 } \nonumber$$

The moment of inertia is defined as $$\shaded{ I = mr^2 } \nonumber$$

For rotation movements, $$I$$ replaces the mass $$m$$. The rotational kinetic energy is $$\shaded{ \frac{1}{2}\,I\,\omega^2 } \nonumber$$

A solid with density $$\delta_i$$ rotating about the origin.

A tiny area $$\Delta A$$ with mass $$\Delta m=\delta_i\,\Delta A$$, has a moment of inertia $$\Delta m.r^2=\delta.\Delta A.r^2 \nonumber$$

Consider all the pieces $$\shaded{ I_o=\iint_R r^2\,\delta\,dA } \nonumber$$ where $$r^2=x^2+y^2$$ in $$xy$$-coordinates.

#### Rotation about the $$\ x$$-axis

In the $$xyz$$-space, the distance to the $$x$$-axis is $$|y|$$. Solid spinning around $$x$$-axis

Moment of inertia for a solid with density $$\delta$$ rotaring about the $$x$$-axis $$\shaded{ I_x=\iint_R y^\,\delta\,dA } \nonumber$$

#### Examples

##### One

Disk of radius $$a$$ with uniform density $$\delta=1$$ spinning around its center. What is the moment of inertia?

What is $$r^2$$ for any point inside $$R$$ in this formula? $$I_o=\iint_R r^2\,\delta\,dA \nonumber$$

Using polar coordinates, $$r$$ will go from $$0$$ to $$a$$ and $$dA=r\,dr\,d\theta$$ \begin{align*} I_o &= \iint_R r^2.1.dA \\ &= \int_0^{2\pi} \underline{\int_a^a r^2 r\,dr}\,d\theta \\ &= \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_{r=0}^a\,d\theta = \int_0^{2\pi} \frac{a^4}{4}\,d\theta \\ &= \frac{a^2}{4}\Big[\theta\Big]_0^{2\pi} = \frac{1}{2}\pi a^4 \end{align*} \nonumber

##### Two

How much harder is it to spin this disk around a point on its circumference?

The inertia \begin{align*} I_o & =\iint r^2\,dA \\ &= \int_{-\pi/2}^{\pi/2} \underline{\int_0^{2a\cos\theta} r^2 r\,dr}\,d\theta \\ \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_0^{2a\cos\theta} r^2 r\,dr &=\left[\frac{r^4}{4}\right]_{r=0}^{2a\cos\theta} \\ &= 4a^4\cos^4\theta \end{align*} \nonumber

Evalutate the outer integral \begin{align*} I_o & =\iint r^2\,dA \\ &= \int_{-\pi/2}^{\pi/2} 4a^4\cos^4\theta\,d\theta = \dots = \frac{3}{2}\pi a^4 \end{align*} \nonumber

It is three times harder to spin a Frisbee about a point on a circumference than around the center.

## Change of variables

We change variables, when it simplifies the integrand or bounds, so it becomes easier to compute the double integral.

### Examples

#### One

Determine the area of an ellipse with semi-axes $$a$$ and $$b$$. $$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1 \nonumber$$

The double integral for the area $$\rm{Area} = \iint_{\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2\lt 1} dx\,dy \nonumber$$

Use substitution to make it look more like a circle \left. \begin{array}{c} \text{set }\frac{x}{a}=u \Rightarrow du = \frac{1}{a}dx \\ \text{set }\frac{y}{a}=v \Rightarrow dv = \frac{1}{b}dy \end{array} \right\} \\ \begin{align*} \Rightarrow du\,dv &= \frac{1}{ab}dx\,dy \\ \Rightarrow dx\,dy &= ab\,du\,dv \end{align*} \nonumber

Substitute it back in the double integral \begin{align*} \rm{Area} &= \iint_{u^+v^2\lt 1} ab\,du\,dv \\ &= ab\underbrace{\iint_{u^+v^2\lt 1} du\,dv}_{\text{area of unit disk}} = a\,b\,\pi \end{align*} \nonumber

#### Two

To simply integrand or bounds, we set a change of variables as \left\{ \begin{align*} u &= 3x-2y \\ v &= x+y \end{align*} \right. \nonumber

What is the relation between $$dA=dx\,dy$$ and $$dA’=du\,dv$$? $$\Delta x, \Delta y$$ $$\Delta u, \Delta v$$

The linear transformation changes it to a parallelogram. Because of the linear change of variables, the area scaling factor doesn’t depend on the choice of rectangle. So let’s take the simplest rectangle, the unit square. Simplest rectangle in $$xy$$

Applying the transformation to the corners Simplest rectangle in $$uv$$

The area $$A’$$ is the determinant of the two vectors from the origin $$A’ = \left| \begin{array}{rr} 3 & 1 \\ -2 & 1 \end{array} \right| = 3+2=5 \nonumber$$

For any other rectangle, area is also multiplied by $$5$$ \begin{align*} dA’ &= 5\,dA \\ \Rightarrow du\,dv &= 5\,dx\,dy \\ \Rightarrow \iint\ldots\,dx\,dy &= \iint\ldots\,\frac{1}{5}du\,dv \end{align*} \nonumber

### Jacobian

Changing variables to $$u,v$$ means \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \left\{ \begin{align*} u = u(x,y) &\Rightarrow \Delta u\approx \pdv{u}{x}\Delta x+\pdv{u}{y}\Delta y = u_x\Delta x+u_y\Delta y \\ v = v(x,y) &\Rightarrow \Delta v\approx \pdv{v}{x}\Delta x+\pdv{v}{y}\Delta y = v_x\Delta x + v_y\Delta y \end{align*} \right. \nonumber

In matrix form $$\left[ \begin{array}{c} \Delta u \\ \Delta v \end{array} \right] \approx \left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right] \left[ \begin{array}{c} \Delta x \\ \Delta y \end{array} \right] \nonumber$$

A small rectangle in $$xy$$-coordinates corresponds to a small parallelogram in $$uv$$-coordinates. The sides of the parallelogram from $$(0,0)$$, are the vectors $$\left\langle\Delta x,0\right\rangle$$ and $$\left\langle 0,\Delta y\right\rangle$$ \left\{ \begin{align*} \left\langle\Delta x,0\right\rangle \rightarrow \left\langle\Delta u,\Delta v\right\rangle &\approx \left\langle u_x\Delta x, v_x\Delta x\right\rangle \\ \left\langle 0,\Delta y\right\rangle \rightarrow \left\langle\Delta u,\Delta v\right\rangle &\approx \left\langle u_y\Delta y, v_y\Delta y\right\rangle \end{align*} \right. \nonumber

The area $$\rm{Area}’$$ of the parallelogram is the determinant $$\text{Area}’ = \rm{det} \left( \left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right] \right) \Delta x\,\Delta y \nonumber$$

When you have a general change of variables, $$du\,dv$$ versus $$dx\,dy$$ is given by the determinant of the matrix of partial derivatives. $$\rm{det}\left( \left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right] \right) \nonumber$$

The definition of Jacobian just means the ratio between $$du\,dv$$ and $$dx\,dy$$. (Not a partial derivative.) Here the vertical bars stand for determinant.

$$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \shaded{ J = \pdv{(u,v)}{(x,y)} = \left| \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right| } \nonumber$$

Then, because area is always positive $$\shaded{ du\,dv = |J|\,dx\,dy = \left|\pdv{(u,v)}{(x,y)}\right|\,dx\,dy } \nonumber$$

### Examples

#### One

Switching to polar coordinates \begin{align*} x &= r\cos\theta \\ y &= r\sin\theta \end{align*} \nonumber

The Jacobian \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \pdv{(x,y)}{(r,\theta)} &= \left| \begin{array}{cc} x_r & x_\theta \\ y_r & u_\theta \end{array} \right| \\ &= \left| \begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array} \right| \\ &= r\cos^2\theta – (-r\sin^2\theta) \\ &= r(\cos^2\theta + \sin^2\theta) = r \end{align*}

Not a constant, but a function of $$r$$, so \shaded{ \begin{align*} dx\,dy &= |r|\,dr\,d\theta \\ &= r\,dr\,d\theta \end{align*} } \nonumber

Remark: you can compute the one that easier to compute, because they are the inverse of each other. $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \pdv{(u,v)}{(x,y)} \cdot \pdv{(x,y)}{(u,v)} = 1 \nonumber$$

#### Two

Compute $$\int_0^1\int_0^1 x^2y\,dx\,dy \nonumber$$

using change of variables to \left\{ \begin{align*} u &= x \\ v &= xy \end{align*} \right. \nonumber

Step 1: Find the area element using the Jacobian \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \pdv{(x,y)}{(r,\theta)} &= \left| \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right| \\ &= \left| \begin{array}{cc} 1 & 0 \\ y & x \end{array} \right| = x \end{align*} With $$x$$ positive in the region \begin{align*} du\,dv &= |x|\,dx\,dy \\ &= x\,dx\,dy \end{align*} \nonumber

Step 2: Express the integrand in terms of $$u,v$$ \begin{align*} x^2y\,dx\,dy &= x^2y\,\frac{1}{x}\,du\,dv = xy\,du\,dv \\ &= u\frac{v}{u}\,du\,dv = v\,du\,dv \end{align*} \nonumber Compute (or $$dv\,du$$) $$\iint_\ldots v\,du\,dv \nonumber$$

Step 3: Find the bounds for $$u,v$$ in the new integral \begin{align*} \int_\ldots^\ldots \underbrace{\int_\ldots^\ldots v\,du}_{u \text{ changes},\\ v\text{ is constant}}\,dv \end{align*} \nonumber $$v=\rm{constant} \rightarrow xy=\rm{constant} \rightarrow y=\frac{\rm{constant}}{x}$$ $$xy$$ and $$uv$$-coordinates
What is the value of $$u$$ when we enter the region from the top, where $$y=1$$? \begin{align*} y &=1 \\ \Rightarrow y &=\frac{v}{u}=1 \\ \Rightarrow u &= v \end{align*} What is the value of $$u$$ when we exit the region, where $$x=1$$? \begin{align*} x &=1 \\ \Rightarrow u &= 1 \end{align*} The smallest value of $$(x,y)$$ is $$(0,0)$$, what corresponds to $$v=0$$. The largest value of $$(x,y)$$ is $$(1,1)$$, what corresponds to $$v=1$$.

Step 4: The double integral follows as \begin{align*} \int_0^1 \int_v^1 v\,du\,dv \end{align*} \nonumber $$\nonumber$$

How could we have found the bounds easier? Draw the picture the $$uv$$-coordinates. $$xy$$-coordinates $$uv$$-coordinates