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, 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


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 $$ \shaded{ \iint_R f(x,y)\,dA = \int_{x_{min}}^{x_{max}} \left[ \int_{y_{min}(x)}^{y_{max}(x)} f(x,y)\,dy \right] dx } \nonumber $$



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


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


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


\(z=f(x,y)\) 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.



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


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.


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


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.

Simply connected regions
Not simply connected regions

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 $$ \shaded{ \int_{\theta_{min}}^{\theta_{max}} \int_{r_{min}}^{r_{max}} f(r,\theta)\,r\,dr\,d\theta } \nonumber $$



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


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).

Linear motion
Circular motion

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 $$ \shaded{ \frac{1}{2}\,I\,\omega^2 } \nonumber $$

Rotation about the origin

A solid with density \(\delta_i\) rotating about the origin.

Solid rotating around 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 $$ \shaded{ I_x=\iint_R y^\,\delta\,dA } \nonumber $$



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

Disk spinning around origin

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 \(dA=r\,dr\,d\theta\) $$ \begin{align*} I_o &= \iint_R r^2.1.dA \\ &= \int_0^{2\pi} \underline{\int_a^a r^2 r\,dr}\,d\theta \\ &= \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_{r=0}^a\,d\theta = \int_0^{2\pi} \frac{a^4}{4}\,d\theta \\ &= \frac{a^2}{4}\Big[\theta\Big]_0^{2\pi} = \frac{1}{2}\pi a^4 \end{align*} \nonumber $$


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

Disk spinning around 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.



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


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 \(5\) $$ \begin{align*} dA’ &= 5\,dA \\ \Rightarrow du\,dv &= 5\,dx\,dy \\ \Rightarrow \iint\ldots\,dx\,dy &= \iint\ldots\,\frac{1}{5}du\,dv \end{align*} \nonumber $$


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

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.

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

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



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

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

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. $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \pdv{(u,v)}{(x,y)} \cdot \pdv{(x,y)}{(u,v)} = 1 \nonumber $$


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 $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \pdv{(x,y)}{(r,\theta)} &= \left| \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right| \\ &= \left| \begin{array}{cc} 1 & 0 \\ y & x \end{array} \right| = x \end{align*} $$ 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.


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