1. ## circle transformation

Hi,

I have been given the following transformation:

$x=u+2v+1$
$y=2u+v+1$

which transforms the circle $u^2+v^2<=1$ to the region R in the XY field.

I am requested to find the Volume of R.

I tried using a double integral with polar coordinates, but couldn't exactly understand how i am supposed to use the transformation.

Can someone help me understand what I should be doing here ?

2. Originally Posted by meirgold
Hi,

I have been given the following transformation:

$x=u+2v+1$
$y=2u+v+1$

which transforms the circle $u^2+v^2<=1$ to the region R in the XY field.

I am requested to find the Volume of R.

I tried using a double integral with polar coordinates, but couldn't exactly understand how i am supposed to use the transformation.

Can someone help me understand what I should be doing here ?
the Jacobian of the transformation is: $J=\det \begin{pmatrix} 1 & 2 \\ \ 2 & 1 \end{pmatrix} =-3.$ so: the area of $R=\int \int_{u^2 + v^2 \leq 1} |J| \ du \ dv=3\pi. \ \ \ \square$

3. ## not sure i understood

Hi,

thanks for the reply, but i'm not sure i understood 2 matters:

1. What are the numbers which you used for the double integral ? I noticed that you used: $u^2+v^2<=1$, but what exact values did you extract from that formula ?

2. I need to find the volume of R and not the area. Is there a way to find the volume from the area ?

Thanks

4. Originally Posted by meirgold
Hi,

I have been given the following transformation:

$x=u+2v+1$
$y=2u+v+1$

which transforms the circle $u^2+v^2<=1$ to the region R in the XY field.

I am requested to find the Volume of R.

I tried using a double integral with polar coordinates, but couldn't exactly understand how i am supposed to use the transformation.

Can someone help me understand what I should be doing here ?
Huh? Are you sure that you are asked to find the volume of R? Then I'm afraid that it would be 0 since there are only 2 dimensions. Are you sure that you posted the question correctly?

5. ## oooops

sorry, you are correct - i was refering to the area.

Can you explain the answer to my first question ?

Thanks

6. Originally Posted by meirgold
Hi,

thanks for the reply, but i'm not sure i understood 2 matters:

1. What are the numbers which you used for the double integral ? I noticed that you used: $u^2+v^2<=1$, but what exact values did you extract from that formula ?

2. I need to find the volume of R and not the area. Is there a way to find the volume from the area ?

Thanks
You're given $u^2+v^2<=1$ in the question so there's no mystery there. It's a unit circle so the double integral clearly represents three times the area of a unit circle, so there's no mystery there either.

I don't understand exactly where your trouble is and what exact values you're referring to ......

7. I'll try to be more clear:

Each of the integrals (within the double integral) should have a lower limit and an upper limit.
I understand that somehow I'm supposed to derive them from the equation: $u^2+v^2<=1$ but am not sure how to do this.

In addition, if the answer is $3\pi$, than should I be using polar coordinates ? If so, shouldn't the Jacobian be equal to the circle radius (which is r=1) ?

8. Originally Posted by meirgold
I'll try to be more clear:

Each of the integrals (within the double integral) should have a lower limit and an upper limit.
I understand that somehow I'm supposed to derive them from the equation: $u^2+v^2<=1$ but am not sure how to do this.

In addition, if the answer is $3\pi$, than should I be using polar coordinates ? If so, shouldn't the Jacobian be equal to the circle radius (which is r=1) ?

You don't need to evaluate any integral. We didn't even convert the circle region to polar coords because we simply know what a circle's area is.

I'll try to explain it step by step.

You first have a circle, $u^2 + v^2 \leq 1$. Call this region D. The area of this circle is,

$\int \int_D~du~dv$.

Transform this to x-y coordinates, call the new region R. This transformation will give,

$\int \int_R |J|~dx~dy$.

So, $\int \int_D~du~dv = \int \int_R |J|~dx~dy$.

Remember, $\int \int_D~du~dv$ represents the area of a homogenous unit circle (a circle of radius 1). We know that it's $\pi r^2 = \pi$.

$\pi = \int \int_R |J|~dx~dy$.

$|J|$ will be a constant (this is not a general rule but a special case). So we can write,

$\underbrace{\int \int_R ~dx~dy}_{\text{area of R}}=\frac{\pi}{|J|}$

Note that this Jacobian is different than NonCommAlg's, because I transformed from x-y to u-v while he did it the other way.

9. Thanks.
It's all clear now.