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Change in variables of double integrals

The question is find

\displaystyle\int \displaystyle\int_{P} x^2+y^2\ dxdy

for the parallelogram bound by the equations

x+y=1

x+y=2

3x+4y=5

and

3x+4y=6

.

What I've done so far is find a sketch and the coordinates of the vertices which are (-2,3), (-1,2), (2,0) and (-1,3).

Basically I've been given a series of double integrals by an academic but I don't think they realised that we're meant to use a change of variable.

\displaystyle\int_{-1}^{-2} \displaystyle\int_{1-x}^{(6-3x)/4}\ x^2+y^2 dxdy

\displaystyle\int_{-1}^{2} \displaystyle\int_{(5-3x)/4}^{(6-3x)/4}\ x^2+y^2 dxdy

+

And there is also another integral but latex won't let me type it up because its dangerous apparently... either way this approach is futile because the integrals give expressions which appear far too long.

For a change in variable, what would I put as u and v? I'm trying

u=x+y

and

v=3x+4y

but I'm not sure what to do next. Would it result in 1 double integral opposed to 3?

Reply 1

atsruser

Original post by TajwarC

The question is find

\displaystyle\int \displaystyle\int_{P} x^2+y^2\ dxdy

for the parallelogram bound by the equations

x+y=1

x+y=2

3x+4y=5

and

3x+4y=6

.

...

For a change in variable, what would I put as u and v? I'm trying

u=x+y

and

v=3x+4y

but I'm not sure what to do next. Would it result in 1 double integral opposed to 3?

That looks sensible, as it will give you constant limits in

(u,v)

space. These will be easier to integrate with. The Jacobian is also straightforward to calculate. And yes, you will have 1 double integral.

Reply 2

TajwarC

Original post by atsruser

That looks sensible, as it will give you constant limits in

(u,v)

space. These will be easier to integrate with. The Jacobian is also straightforward to calculate. And yes, you will have 1 double integral.

I've calculated the Jacobian as 1 and have this:

\displaystyle\int \displaystyle\int \ (4u-v)^2+(v-3u)^2 dudv

How do I find the limits?

Reply 3

RichE

Original post by TajwarC

I've calculated the Jacobian as 1 and have this:

\displaystyle\int \displaystyle\int \ (4u-v)^2+(v-3u)^2 dudv

How do I find the limits?

In your original post you say what u and v vary between.

Reply 4

TajwarC

Original post by RichE

In your original post you say what u and v vary between.

well

1 \leq u\leq 2

and

5 \leq v\leq 6

which gives:

\displaystyle\int_{5}^{6} \displaystyle\int_{1}^{2} \ (4u-v)^2+(v-3u)^2 dudv

but aren't the inner integral limits meant to be in terms of v and not constants? Or is my actual integrand wrong?

(edited 6 years ago)

Reply 5

RichE

Original post by TajwarC

well

1 \leq u\leq 2

and

5 \leq v\leq 6

which gives:

\displaystyle\int_{5}^{6} \displaystyle\int_{1}^{2} \ (4u-v)^2+(v-3u)^2 dudv

but aren't the inner integral limits meant to be in terms of v and not constants? Or is my actual integrand wrong?

Your limits are functions of v, just constant functions.

What would your limits be if you integrated using x and y over the unit square?

Reply 6

TajwarC

Original post by RichE

Your limits are functions of v, just constant functions.

What would your limits be if you integrated using x and y over the unit square?

So with those limits I got the answer as 7/2.

The limits for an integral for x and y over a region is where most of my confusion lies, I don't know how to find them.

Given