# Getting a double integral over a region

1. May 18, 2013

### Emspak

1. The problem statement, all variables and given/known data
Let B $\in$ ℝ2

the region is bounded by $x^2 + y^2 = 4, \ x^2 + y^2 = 1, \ x^2 = y, \ 2x^2 = y$

evaluate

$$\iint \limits_{B}\frac {2x^2+y^2}{xy}$$

The attempt at a solution

I needed to get the limits of integration. I used the following formula to start with:

$$\iint \limits_{B}f \circ g = \iint \limits_{B}(f \circ g) J (f \circ g)$$

So I let $s = y-x^2, \ t=1-x^2 - y^2$ as those are easy to work with at the start. I do the following:

$\frac{\partial (s,t)}{\partial (x,y)}= \begin{vmatrix} -2x & 1\\ -2x & -2y\\ \end{vmatrix}$

The determinant gets me $4xy +2x$ and the inverse of that is $\frac {1}{4xy +2x}$

So I put it all together thus:
$$\iint \limits_{B}\frac {2x^2+y^2}{xy}\frac {1}{4xy +2x} = \iint \limits_{B}\frac {2x^2+y^2}{4x^2y^2+2x^2y}$$

To get the integration limits I look at where the various curves meet. In the case of y=x2 and x2+y2=1 it's at (-1/√2, 1/√2) and the parabola meets the other circle at (-√2, √2). The y=2x2 curve meets the circles at (-1/√3, √2/√3) and (-2/√3, 2√2/√3).

so the limits of x are -1/√3 to -√2 and y is from 1/√2 to 2√2/√3.

That gives me

$$\int\limits_{-1/\sqrt{3}}^{-\sqrt{2}}\int\limits_{1/\sqrt{2}}^{2\sqrt{2}/\sqrt{3}} \frac {2x^2+y^2}{4x^2y^2+2x^2y}\,dy\,dx$$

So my question is, what I did wrong here :-) since odds are there is something.

2. May 18, 2013

### Emspak

I just realized i think I should have set the y-limits as

y=x^2 and y=2x^2

rather than the numbers I had -- yes?

3. May 19, 2013

### haruspex

I don't understand what you've done. You introduced s and t and calculated the Jacobian, so next I expected to see an integral written entirely in terms of s, t, ds, dt. But you continued with x, y, dx, dy, producing an integral which is clearly different from the one you started with.
My inclination would be just to convert to polar. The tricky bit will be determining the range for theta as a function of r.

4. May 19, 2013

### Emspak

Hmmm. Let's say I convert to polar coordinates. i get this, them using x=rcosθ and y=rsinθ:

$$\iint \limits_{B}\frac {2r\cos^2\theta+r\sin^2\theta}{r\cos\theta r\sin\theta}$$
or
$$\iint \limits_{B}\frac {2r\cos^2\theta+r\sin^2\theta}{r^2\cos\theta\sin\theta}\,dr\,d\theta = 1$$

To do the limits of integration I can then use the parabola and the circle, where instead of $y = x^2, \ 1=x^2 + y^2$ I can use $r\sin\theta = r^2\cos\theta$ and $r\cos^2\theta+r\sin^2\theta = 1$

The Jacobian would then look like this:
$\frac{\partial (r,θ)}{\partial (θ,r)}= \begin{vmatrix} \sec^2\theta & \frac {r}{1 + r^2}\\ 0 & \cos^2\theta+\sin^2\theta\\ \end{vmatrix}$

How are we doing so far?

5. May 19, 2013

### SteamKing

Staff Emeritus
You haven't done the conversion to polar coordinates correctly. If x = r*cos (theta), what is x^2 in polar coordinates? Ditto y.

6. May 19, 2013

### Emspak

Dammit, you're right. It should be

$$\iint \limits_{B}\frac {2r^2\cos^2\theta+r^2\sin^2\theta}{r^2\cos\theta\sin\theta}$$

So geting back to the limits of integration, $y = x^2, \ 1=x^2 + y^2$ I can use $r\sin\theta = r^2\cos\theta$ and $r\cos^2\theta+r\sin^2\theta = 1$ and r goes to 1 in the second equation.

The Jacobian would then look like this:
$\frac{\partial (r,θ)}{\partial (θ,r)}= \begin{vmatrix} r\sec^2\theta & \frac {r}{1 + r^2}\\ 0 & 0\\ \end{vmatrix}$ with a determinant of 0, though, no?

7. May 19, 2013

### SammyS

Staff Emeritus
No.

In going from Cartesian coordinates: x, y to polar: r, θ the Jacobian is the well known r .

I.e. $\displaystyle \ \int\int_B f(x,y)dxdy → \int\int_B f(r\cos\theta,\,r\sin\theta)r\,drdθ$

As you should know, the differentials which are included as part of the integration symbolism are very important. You have dropped them altogether.

Frankly, I'm not convinced that polar coordinates are the way to go. They very well might work. After all haruspex has an excellent reputation as a Homework Helper here at PF. But I'm having difficulty working with the boundaries, y = 2x2 and y = x2 and converting them to workable integration limits for θ .

I suggest sketching a graph of region B. When I did that I noticed that region B consists of two two separate disjoint regions. One in Quadrant I, the other in Quadrant II. Is region B limited to Quadrant I ?

Do you have some additional instructions regarding this problem? -- instructions you haven't shared with us?

Going back to your initial attempt: It's usually the boundaries that you want to convert from curves to straight line segments. Of course, the resulting integrand needs to be reasonable to work with.

8. May 19, 2013

### LCKurtz

When I first saw this problem I was tempted to ask the OP whether it was copied correctly. The natural change of variables would be $u = x^2 + y^2,\, v = \frac y {x^2}$. That gives a rectangular region in the $uv$ plane: $1\le u\le 4,\, 1 \le v \le 2$. But after changing the variables it is still a mess in the integrand, making me wonder if it was posted correctly.

9. May 19, 2013

### Emspak

I just realized I forgot a very important bit about which quadrant it was in: the original problem says it is quadrant 2.

I also realized that I posted the original problem slightly wrong -- and the "real" version is a lot simpler. The integrand to be evaluated is:

$$\iint \limits_{B}\frac {2x^2+4y^2}{xy}$$

Either way, I screwed up going to polar coordinates. Argh. But if you don't, let's look at this another way: would the limits of integration be, in the x-coordinate (the inner integral) something like x=√y and x=√y/√2?

(I'm sorry I posted this wrong and wasted people's time)

10. May 19, 2013

### Emspak

OK, I corrected the statement of the problem. You said the natural change of variables would be as above.

So what's the next step? Plainly I have zero idea. What you did gives me integration limits, right?

11. May 19, 2013

### LCKurtz

Yes, do you see why? What happens when you calculate$$J=\frac{\partial(u,v)}{\partial(x,y)}$$ and put $\frac 1 J$ in the integrand to change the variables to $u$ and $v$?

12. May 19, 2013

### SammyS

Staff Emeritus
You continue to leave the differentials, in this case, dxdy, out of your integral expressions .

It's $\ \displaystyle \iint \limits_{B}\frac {2x^2+4y^2}{xy}\textbf{dx dy} \ .$

13. May 19, 2013

### Emspak

$$J=\frac{\partial(u,v)}{\partial(x,y)}$$

gets me

$\frac{\partial (s,t)}{\partial (x,y)}= \begin{vmatrix} 2x & 2y\\ \frac{y}{3x^2} & \frac{1}{x^2}\\ \end{vmatrix}$

and the determinant is $$\frac{2}{x} - \frac{2y^2}{3x^2}$$ or $$\frac{6x-2y^2}{3x^2}$$

and the inverse is $$\frac{3x^2}{6x-2y^2}$$

am I multiplying this by the original integrand?

14. May 19, 2013

### LCKurtz

Write down your substitution: $s = x^2+y^2,\, t = \frac y {x^2}$ so we can follow your work.

Nope. Get it right.

You might want to read what the original change of variables theorem says to understand what to do next.

15. May 19, 2013

### Emspak

Which are you referring to here? I am not trying to be obtuse. I really just want to now what I did wrong in getting that matrix. If I knew I wouldn't ask, you know? But at this point I am just lost. Taking a derivative of u and v relative to x and y (as per what you used in your substitution) got me the matrix above, no? Is that what is wrong? THe derivative of t w/r/t y is 1/x2, isn't it?

(I did look at the original theorem. That isn't helping me much right now).

At this point I am just lost as all get out.

16. May 19, 2013

### SammyS

Staff Emeritus
If $\displaystyle \ t=\frac{y}{x^2}\,,\$ then $\displaystyle \ \frac{\partial t}{\partial x} = \frac{-2y}{x^3}$

17. May 19, 2013

### Emspak

OK, thanks, SammyS, I realize what I did wrong there, a least.
So my matrix should look like:

$\frac{\partial (s,t)}{\partial (x,y)}= \begin{vmatrix} 2x & 2y\\ \frac{-2y}{x^3} & \frac{1}{x^2}\\ \end{vmatrix}$

correct? And the determinant should then be $$\frac{2}{x} - \frac{-4y^2}{x^3}$$ which can be reduced to $$\frac{2x^2+4y^2}{x^3}$$ and the inverse of that is $$\frac{x^3}{2x^2+4y^2}$$

have I at least got that right?

and if so, what I am trying to understand is what the next step is. I have an inverse of the Jacobian matrix, yes?

$$\iint \limits_{B}\frac {2x^2+4y^2}{xy}\frac{x^3}{2x^2+4y^2}\,dx \ dy$$

which simplifies down to

$$\iint \limits_{B}\frac {x^2}{y}\,dx \ dy$$ and the final integrand should look like this:

$$\int \limits_1^4 \int \limits_{x^2}^{2x^2} \frac {x^2}{y}\,dx \ dy$$

I THINK I am starting to get this but I am not entirely sure, and reading the change of variables theorem wasn't terrifically enlightening. Because the change of variables formula actually looks rather different from what I have done here.

18. May 19, 2013

### SammyS

Staff Emeritus
Yes, that's right -- if you write it as a matrix, not a determinant.

Not exactly. You could do that. It's a lot of work. But you only need the determinant of the inverse of this matrix. That's simply the multiplicative inverse of the determinant of the above Jacobian matrix.

At this point the variables of integration have been changed to s and t, so you should have
$\displaystyle \iint \limits_{B}\frac {2x^2+4y^2}{xy}\frac{x^3}{2x^2+4y^2}\,ds \, dt$​
with the understanding that the integrand is to be in terms of the variables on integration, s and t .

the rest id wrong, since it's not integrated over s and t.
For that last integraand: What is $\displaystyle \ \frac {x^2}{y}\$ in terms of s and t ?

19. May 19, 2013

### Emspak

well, $$t=\frac{y}{x^2}\$$ and $$s= x^2 + y^2$$

and since $$x^2 = \frac{y}{t}$$

I end up with

$$\int \limits_1^4 \int \limits_{x^2}^{2x^2} \frac {1}{t}\,ds \ dt$$

and substituting in the integrand limits

$$\int \limits_1^4 \int \limits_{\frac{y}{t}}^{2\frac{y}{t}} \frac {1}{t}\,ds \ dt$$

so far so good?

20. May 19, 2013

### SammyS

Staff Emeritus
The limits of integration are incorrect.

Look back to see why LCKurtz suggested this change of variables in the first place.