# Setting up an integral over a parallelogram

edited April 8

10 points

Your mission in this problem is to set up an integral over a your own, personal parallelogram. First, get your own, personal parallelogram $P$ from the Assignments section under the "Change of variables" section of Tuesday, April 7 on our Online Presentations page. Once you've got it, set up
$$\iint_P f(x,y) \, dA.$$
as a single iterated integral involving the variables $s$ and $t$. Note that this is an integration formula for a generic function. Thus, your final answer should look something like:
$$\int_c^d \int_a^b f(x(s,t), y(s,t)) \color{red}? \, ds \, dt,$$
where $[a,b]\times[c,d]$ defines a rectangle in the $st$-plane, $x(s,t)$and $y(s,t)$ are functions, and $\color{red}?$ is the adjustment factor arising from the Jacobian.

If you'd like to check that your answer is reasonable, you might plot the parallelogram determined by your functions using our parametric region plotter.

## Comments

• edited April 8

My bounds were: 5x-y=0, 5x-y=2, 3y-x=0, 3y-x=2
Set s=5x-y and t=3y-x

Gives 0<s<2 and 0<t<2

Solve for x and y
5x-s=y, (t+x)/3=y and 3y-t=x, (s+y)/5=x

Set equal to find x and y in terms of s and t
5x-s=(t+x)/3 -> x=(t+3s)/14
(s+y)/5=3y-t -> y=(s+5t)/14

Take respective derivatives.
dx/ds=3/14, dx/dt=1/14, dy/ds=1/14, dy/dt=5/14

Find the Jacobian
|[3/14,1/14],[1/14,5/14]|=1/14

Plug everything in to get this mess
int_0^2int_0^2f((3s+t)/14,(s+5t)/14)1/14dsdt

• edited April 8

My parallelogram is bounded by 4x-y=0, 4x-y=3, 5y-x=0, and 5y-x=5

s=4x-y and t= 5y-x
So, 0<s<3 and 0<t<5

Solving for x and y:
y= 4x-s and y=(t+x)/5
x= 5y-t and x=(s+y)/4

Setting equal:
4x-s=(t+x)/5 ------> x= (t+5s)/19
5y-t=(s+y)/4-------> y= (s+4t)/19

Finding the derivatives:
dx/(ds)= 5/19, dx/(dt)= 1/19
dy/(ds)=1/19, dy/(dt)=4/19

Finding the Jacobian:
|[5/19, 1/19],[1/19,4/19]| = 1/19

The integral:
int_0^5int_0^3f((5s+t)/19, (4t+s)/19)(1/19)dsdt

• edited April 9

My parallelogram is bounded by the lines 2x-y=0, 2x-y=4, 4y-x=0, and 4y-x=3

From this, we can find that:

0<s<4 and 0<t<3, and s=2x-y, t=4y-x

Solving for x and y in terms of s and t gives us:

x=(4s+t)/7 and y=(s+2t)/7

The resulting Jacobian is:

|[4/7,1/7],[1/7,2/7]|=8/49-1/49=1/7

As a result, the integral over this parallelogram is:

$$\int_0^3 \int_0^4 f\bigg(\frac{4s+t}{7},\frac{s+2t}{7}\bigg)\bigg(\frac{1}{7}\bigg)dsdt.$$

• edited April 9

My bounds were: $3x-y=0$, $3x-y=3$, $4y-x=o$, and $4y-x=2$.

Set $s=3x-y$ and $t=4y-x$

Give $0<s<3$ and $0<t<2$

Solve for $x$ and $y$

$y=3x-s$, $y=\frac{t+x}{4}$, and $x=\frac{s+y}{3}$, $x=\frac{t+4s}{11}$

Set equal to find $x$ and $y$ in terms of $s$ and $t$

$\frac{s+y}{3} = 4y-t \Rightarrow y= \frac{s+3t}{11}$

$\frac{t+x}{4}=3x-s \Rightarrow x=\frac{t+4s}{11}$

Take respective derivatives

$\frac{dx}{ds}=\frac{4}{11}, \frac{dx}{dt}=\frac{1}{11}, \frac{dy}{ds}=\frac{1}{11}, \frac{dy}{dt}=\frac{3}{11}$

Find Jacobian

$\Bigg| \frac{4}{11} \frac{1}{11}\Bigg|$

$\Bigg| \frac{1}{11} \frac{3}{11} \Bigg| =\frac{1}{11}$

As result the integral is:

$$\int_{0}^{2} \int_{0}^{3} f(\frac{t+4s}{11}, \frac{s+3t}{11}) (\frac{1}{11})dsdt$$

• edited April 9

The information I was given was $3x-y=0$, $3x-y=4$, $3y-x=0$, and $3y-x=2$.

When I set it into terms of s and t, I got $s=3x-y$ and $t=3y-x$.

Given that 0<s<2 and 0<t<4,

solving for x and y I got; $\frac{3s+t}{8}=x$ and $\frac{3t+s}{8}=y$.

The derivatives to find the Jacobian are:
$\frac{dx}{ds}=\frac{3}{8}$, $\frac{dx}{dt}=\frac{1}{8}$, $\frac{dy}{ds}=\frac{1}{8}$, $\frac{dy}{dt}=\frac{3}{8}$.

The Jacobian:
|$\frac{3}{8}$ $\frac{1}{8}$|
|$\frac{1}{8}$ $\frac{3}{8}$| = $\frac{1}{8}$

Thus my integral is:
$$\int^2_0 \int^4_0 f(\frac{3s+t}{8},\frac{3t+s}{8}) \frac{1}{8} dsdt$$

• edited April 9

My bounds were:
4x-y=0, 4x-y+3, 4y-x=0, and 4y-x=2

first I set u=4x-y and v=4y-x such that 0<u<3 and 0<v<2

Then I solved for x and y resulting in:
x=(4u+v)/15 and  y=(4v+u)/15

Then I computed the Jacobian:
 |[4/15,1/15],[1/15,4/15]|=1/15

So my integral became:
\int_0^2 \int_0^3 f(\frac{4u+v}{15},\frac{4v+u}{15}) \frac{1}{15} dudv

• edited April 9

The bounds of my parallelogram are 5x-y=0, 5x-y=5, 2y-x=0, 2y-x=4.

In terms of s and t: s=5x-y and t=2y-x.

Based on this 0<s<5 and 0<t<4.

Solve for x and y: y=5x-s, y=(t+x)/2, x=(s+y)/5, x=2y-t

Set them equal to each other to get them in terms of s and t:
5x-s=(t+x)/2 => x=(t+2s)/9
(s+y)/5=2y-t => y=(s+5t)/9

Take derivatives to find the jacobian:
dx/(ds)=2/9, dx/dt=1/9, dy/(ds) =1/9, dy/dt=5/9

The jacobian would then be:
|[2/9,1/9],[1/9,5/9]|=1/9

And therefore the integral:
int_0^4int_0^5f((t+2s)/9,(s+5t)/9)1/9dsdt

• edited April 13

The bounds of my parallelogram are:

4x-y=0, 4x-y=2, 4y-x=0, 4y-x=5

If I set these equations in terms of s and t, I get:

s=4x-y and t=4y-x

From this, I found that,

0<s<2 and 0<t<5

I then solved the above equation to find x and y:

y=4x-s and x=(s+y)/4
y=(t+x)/4 and x=4y-t

Solving to find x and y, I set each x and y equation equal to each other to get"

(s+y)/4=4y-t => y=(s+4t)/15
(t+x)/4=4x-s => x=(t+4s)/15

Now, using the Jacobian:

dx/(ds)=4/15, dx/(dt)=1/15, dy/(ds)=1/15, dy/(dt)=4/15

|[4/15,1/15],[1/15,4/15]| = 1/15

The integral then is:

int_0^5int_0^2f((t+4s)/15,(s+4t)/15)(1/15)dsdt

• edited April 13

My parallelogram was defined by

5x-y=0, 5x-y=4, 2y-x=0, 2y-x=3

First set these equations to s and t...

s=5x-y and t=2y-x

0<s<4 and 0<t<3

Then solve for x and y...

 x=(t+2s)/9

y=(s+5t)/9

Then find the Jacobian...

dx/(ds) = 2/9 dx/(dt) = 1/9 dy/(ds)= 1/9 dy/(dt) = 5/9

|[2/9,1/9],[1/9,5/9]| = 9/89 = 1/9

The integral is then...

int_0^3int_0^4f((t+2s)/9,(s+5t)/9)(1/9)dsdt

• edited April 14

Parallelogram:
$5x-y=0$, $5x-y=3$, $2y-x=0$, $2y-x=4$

Set equations equal to $s$ and $t$:
$s=5x-y$ $\rightarrow$ $0<s<3$
$t=2y-x=0$ $\rightarrow$ $0<t<4$

Solve for $x$ and $y$ using $s$ and $t$ equations:
$s=5x-y$ $\Rightarrow$ $x=\frac{s+y}{y}$ and $y=5x-s$
$t=2y-x$ $\Rightarrow$ $x=2y-t$ and $y=\frac{t+x}{2}$

Set $x$ and $y$ equations equal to each other:
$\frac{s+y}{5}=2y-t$ $\Rightarrow$ $y=\frac{s+5t}{9}$
$5x-s=\frac{t+x}{2}$ $\Rightarrow$ $x=\frac{t+2s}{9}$

Find the Jacobian:
$\frac{dx}{ds}=\frac{2}{9}$ $\frac{dx}{dt}=\frac{1}{9}$
$\frac{dy}{ds}=\frac{1}{9}$ $\frac{dy}{dt}=\frac{5}{9}$
$\frac{d(x,y)}{d(s,t)}=(\frac{2}{9})(\frac{5}{9})-(\frac{1}{9})(\frac{1}{9})=\frac{10}{81}-\frac{1}{81}=\frac{1}{9}$

The integral is:
$\int_{0}^{3}\int_{0}^{4}[f(\frac{t+2s}{9}, \frac{s+5t}{9})] dsdt$

• edited April 14

My parallelogram was bounded by the lines
3x-y=0,3x-y=2,2y-x=0,2y-x=5
This tells me my bounds are
0<s<2
0<t<5
Write these equations in x and y as s and t
s=3x-y => x=(s-y)/3 y=3x-s
t=2y-x => x=2y-t y=(t-x)/2
Then solve for x and y by setting them equal to each other
(s-y)/3=2y-t
3x-s=(t-x)/2
x=(t-2s)/7
y=(3t+s)/7
Then solve Jacobian
dx/(ds)=(-2)/7 dx/(dt)=1/7 dy/(ds)=1/7 dy/(dt)=3/7
|[-2/7,1/7],[1/7,3/7]| = 1/7
Integral is
int_0^2int_0^5f((t-2s)/7,(3t+s)/7)(1/7)dsdt

• edited April 14

My parallelogram was bound by the lines
3x-y=0, 3x-y=5, 4y-x=0, 4y-x=4

The s and t bounds are
0<s<5, 0<t<4

My s and t equations are as follows
s=3x-y => y=3x-s

t=4y-x => x=4y-t

Solving for x and y in terms of s and t I get
(s+y)/3=4y-t => y=(s+3t)/11

(t+x)/4=3x-s => x=(t+4s)/11

My Jacobian will be
dx/(ds) =4/11, dx/dt =1/11

dy/(ds) =1/11, dy/dt =3/11

|[4/11,1/11],[1/11,3/11]|=1/11

Then, my integral will be
int_0^4int_0^5f((t+4s)/11,(s+3t)/11))(1/11)dsdt

• edited April 14

The bounds given are:

3x−y=0, 3 x-y=23x−y=2, 5 y-x=05y−x=0, and 5 y-x=25y−x=2

Meaning that 0<=s<=2 and 0<=t<=2

which are our bounds of integration.

It also means that s=3x+y and t=5y-x

which means y=(s+3t)/14 and x=(5s+t)/14

and the partial derivatives are dx/ds=5/14 , dx/dt=1/14 , dy/ds=1/14 , dy/dt=3/14

Giving our Jacobian of |[5/14,1/14],[1/14,3/14]| equaling 1/14

making the iterated integral:

int_0^2 int_0^2f((5s+t)/14,(3t+s)/14)(1/14)dsdt

• edited April 14

The bounds for my parallelogram are:
4x−y=0, 4 x-y=4, 5 y-x=0, and 5y−x=5

Setting these equal to s and t gives the following bounds:
0<s<4
0<t<5

Solving for x and y gives:
x=\frac{5s+t}{19} and y=\frac{4t+s}{19}

The Jacobian is:
|[5/19,1/19],[1/19,4/19]|
\frac{d(x,y)}{d(s,t)}=\frac{1}{19}

The integral is:
\frac{1}{19}\int_{0}^{5}\int_{0}^{4}f(\frac{5s+t}{19},\frac{4t+s}{19})dsdt

• edited April 14

My parallelogram is bounded by the lines 4x−y=0, 4x−y=3, 4y−x=0, and 4 y-x=5
Setting s=4x−y,t=4y−x
We get 0<s<3 and 0<t<5
Solving for x and y in terms of s and t we get
x=\frac{4s+t}{15} and y=\frac{4t+s}{15}
The Jacobian is
|[4/15,1/15],[1/15,4/15]|=16/225-1/225=15/225=1/15
Which results in the integral
\int_{0}^{5}\int_{0}^{3}f(\frac{4s+t}{15},\frac{4t+s}{15})(\frac{1}{15})dsdt

• edited April 15

My parallelogram is bound by:

4x-y=0
4x-y=3
3y-x=0
3y-x=3

which gives us:

s=4x-y and t=3y-x

Meaning that:

0<s<3
0<t<3

So we solve for x and y, giving us:

y=4x-s -----> (t+x)/3=y
3y-t=x -----> (y=s)/4=x

3y-t=(y+s)/4 -----> y=(s+4t)/11
4x-s=(t+x)/3 -----> x=(t+3s)/11

dx/ds=3/11
dx/dt=1/11
dy/ds=1/11
dy/dt=4/11

The Jacobian is:

| 3/11 1/11 |
|\ 1/11 4/11 |

Which gives us

12/121-1/121=11/121=1/11

And the integral is:

\int_0^3\int_0^3((t+3s)/11,(3+4t)/11)(1/11)\ dsdt

• edited April 15

My parallelogram is bounded by the lines:

$3x-y=0, 3x-y=5, 3y-x=0, 3y-x=4.$

We set $s=3x-y$ and $y=3y-x$ which gives us the bounds $0<x<5$ and $0<t<4.$

We then solve for $x$ and $y$:

$x=\frac{s+y}{3}$, $x=3y=t$, $y=3x-s$, $y=\frac{t+x}{3}$.

We set these equal to give us x and y in terms of s and t:

$\frac{s+y}{3}=3y-t\Longrightarrow y=\frac{s+3t}{8}$ and $\frac{t+x}{3}=3x-s\Longrightarrow x=\frac{t+3s}{8}$.

The Jacobian is:

$\frac{dx}{ds}=\frac{3}{8}$, $\frac{dx}{dt}=\frac{1}{8}$,$\frac{dy}{ds}=\frac{1}{8}$,$\frac{dy}{dt}=\frac{3}{8}$.

$\frac{d(x,y)}{d(s,t)} = \frac{1}{8}$.

This gives us the integral:

$$\int_{0}^{4}\int_{0}^{5}f\bigl(\frac{t+3s}{8},\frac{s+3t}{8}\bigl)\bigl(\frac{1}{8}\bigl)dsdt.$$

• edited May 5

My parallelogram is bounded by the lines

2x-y=0, 2x-y=2,
2y-x=0, 2y-x=3

If we set s=2x-y and t=2x-y
Then we find our bounds: s:[0,2] and t:[0,3]

Then solve for x from both equations and set them equal to each other to find y.

x=(s+y)/2 and x=2y-t

y=(s+2t)/3

Then solve for y from both equations and set them equal to each other to find x.

y=2x-s and y=(t+x)/2

x=(t+2s)/3

Next, find the ~Jacobian~

|[2/3,1/3],[1/3,2/3]|=4/9-1/9=1/3

As a result, the integral over this parallelogram is:

\int_0^2 \int_0^3 f(\frac{2s+t}{3},\frac{s+2t}{3})(\frac{1}{3})dsdt.

• edited April 15

My random parallelogram was bounded by $5x-y=0, 5x-y=4, 3y-x=0,$ and $3y-x=5$. Setting to $s$ and $t$, we get $s=5x-y$ and $t=3y-x$, which gives us the bounds $0 < s < 4$, and $0 < t < 5$.

Solving for $x$ and $y$, respectively, we get

$$x=\frac{y+s}{5}, x=3y+t$$ and $$y=5x-s, y=\frac{x+t}{3}$$

Now we can set our equalities, we can find $x$ and $y$ in terms of $s$ and $t$, and we get

$$y = \frac{s+5t}{14}$$ and $$x=\frac{3s+t}{14}.$$

Now we can grab our partial derivatives, such that
$$\frac{\partial x}{\partial s} =\frac{3}{14}, \frac{\partial x}{\partial t} = \frac{1}{14}$$
$$\frac{\partial y}{\partial s} =\frac{1}{14}, \frac{\partial y}{\partial t} = \frac{5}{14}$$

Now with our partials, we can find the Jacobian.
$$\begin{vmatrix} \frac{3}{14}&\frac{1}{14}\\ \frac{1}{14}&\frac{5}{14} \end{vmatrix}$$
And we get $\frac{1}{14}$.

When we put it all together, we get a double integral that looks something like this.
$$\int_0^5\int_0^4 f (\frac{3s+t}{14},\frac{s+5t}{14})(\frac{1}{14})dsdt.$$

• edited April 15

My parallelogram is: 3x-y=0, 3x-y=2, 5y-x=0, 5y-x=3. I'll set s=3x-y and t=5y-x, which gives me the bounds: 0<s<2 and 0<t<3

Solving for x and y gives me: $$x=(\frac{t+5s}{14})$$ and $$y=(\frac{3t+s}{14})$$

The partial derivatives will be:

$$dx/ds = (5/14), dx/dt=(1/14), dy/ds=(1/14), dy/dt=(3/14)$$

The jacobian will be:

$$15/196 - 1/196 = 1/14$$

So, the final integral will be:

\int_0^3\int_0^2 f (\frac{t+5s}{14},\frac{3t+s}{14})*(\frac{1}{14})dsdt

• edited April 16

My parallelogram is bounded by:

3x-y=0, 3x-y=5, 2y-x=0, 2y-x=3

I'll let s=3x-y and t=2y-x which means the bounds for this integral are: 0<s<5 and 0<t<3

Solving s and t for x and y, we find that:

y=3x-s, y=(t+x)/2, x=(y+s)/3, x=2y-t

Taking the y equations and setting them equal to one another and doing the same for the x equations and solving for x and y, we find:

x=(t+2s)/5 y=(3t+s)/5

Taking the partial derivatives for these equations yields:

dx/(ds)=2/5, dx/dt=1/5, dy/(ds)=1/5, dy/dt=3/5

The Jacobian is:

(6/25)-(1/25)=1/5

Therefore, the final integral is:

int_0^3int_0^5((t+2s)/5,(3t+s)/5)(1/5)dsdt

• edited May 5

The bounds of my parallelogram were:
5x-y=0,
5x-y=4,
2y-x=0,
and
2y-x=4.

s=5x-y which means 0>s>4 and
t=2y-x which means 0>t>4.

Solving s and t for x and y gives:

x=1/5y+1/5s,
x=2y-t,
y=5x-s,
and
y=1/2x+1/2t.

Setting the Ys and the Xs equal to each other gives:

x=1/9t+2/9s and
y=5/9t+1/9s.

Next, the Jacobian is computed as follows:

dx/(ds)=2/9
dx/dt=1/9
dy/(ds)=1/9
dy/dt=5/9

|2/9 1/9|
|1/9 5/9|

=10/81-1/81=9/81=1/9

From here, the Integral becomes:

int_0^4int_0^4f((5/9t+1/9s),(1/9t+2/9s))(1/9))dsdt

• I think you guys are missing an $f$ in your answer - do you see why??

• edited May 7

The bounds of my parallelogram were 5x - y = 0, 5x - y = 3, 3y - x = 0 and 3y - x = 4.

Setting to s and t gives the following bounds:

s=5x-y and t= 3y-x

So that means that 0<s<3 and 0<t<4

Solving for x and y gives:

(s+y)/5=3y-t => y=(s+5t)/14

(t+x)/3=5x-s => x=(t+3s)/14

The Jacobian is:

|[5/14,1/14],[1/14,3/14]|=1/14

This gives us the integral:

\int_{0}^{4}\int_{0}^{3}f(\frac{t+3s}{14},\frac{s+5t}{14})(\frac{1}{14})dsdt

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