Getting a double integral over a region

[Emspak]

Homework Statement

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}<br /> -2x &amp; 1\\<br /> -2x &amp; -2y\\<br /> \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.

Thanks in advance.

 

[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?

 

[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.

 

[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}<br /> \sec^2\theta &amp; \frac {r}{1 + r^2}\\<br /> 0 &amp; \cos^2\theta+\sin^2\theta\\<br /> \end{vmatrix}

How are we doing so far?

 

[SteamKing]

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

 

[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}<br /> r\sec^2\theta &amp; \frac {r}{1 + r^2}\\<br /> 0 &amp; 0\\<br /> \end{vmatrix} with a determinant of 0, though, no?

 

[SammyS]

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}<br /> r\sec^2\theta &amp; \frac {r}{1 + r^2}\\<br /> 0 &amp; 0\\<br /> \end{vmatrix} with a determinant of 0, though, no?

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.

 

[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.

 

[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)

 

[Emspak]

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.

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?

 

[LCKurtz]

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?

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 $v$?

 

[SammyS]

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}

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

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

 

[Emspak]

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

gets me

\frac{\partial (s,t)}{\partial (x,y)}= \begin{vmatrix}<br /> 2x &amp; 2y\\<br /> \frac{y}{3x^2} &amp; \frac{1}{x^2}\\<br /> \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?

 

[LCKurtz]

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

$$
J=\frac{\partial(s,t)}{\partial(x,y)}$$

gets me

\frac{\partial (s,t)}{\partial (x,y)}= \begin{vmatrix}<br /> 2x &amp; 2y\\<br /> \frac{y}{3x^2} &amp; \frac{1}{x^2}\\<br /> \end{vmatrix}

Nope. Get it right.

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?

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

 

[Emspak]

Nope. Get it right.

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.

 

[SammyS]

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

gets me

\frac{\partial (s,t)}{\partial (x,y)}= \begin{vmatrix}<br /> 2x &amp; 2y\\<br /> \frac{y}{3x^2} &amp; \frac{1}{x^2}\\<br /> \end{vmatrix}
...

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

 

[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}<br /> 2x &amp; 2y\\<br /> \frac{-2y}{x^3} &amp; \frac{1}{x^2}\\<br /> \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.

 

[SammyS]

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}<br /> 2x &amp; 2y\\<br /> \frac{-2y}{x^3} &amp; \frac{1}{x^2}\\<br /> \end{vmatrix}

...
$$ \frac{2x^2+4y^2}{x^3}$$ and the inverse of that is $$ \frac{x^3}{2x^2+4y^2}$$

have I at least got that right?

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

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

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.

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

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.

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.

For that last integraand: What is $\displaystyle \ \frac {x^2}{y}\$ in terms of s and t ?

 

[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?

 

[SammyS]

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?

The limits of integration are incorrect.

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

 

[LCKurtz]

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.

You seriously need to study the change of variable theorem. The idea is to change an integral over a region in the $xy$ plane to an integral over a region in the $st$ plane which is easier to evaluate. If you have $x$ and $y$ in terms of $s$ and $t$ then$$
\iint_R f(x,y)dydx = \iint_W f(x(s,t),y(s,t)) |J|dsdt$$where $J$ is the Jacobian$$
\frac{\partial(x,y)}{\partial(s,t)}$$Your new integral will only have $s$ and $t$ variables.

In this problem you have $s$ and $t$ in terms of $x$ and $y$ and the Jacobian you have calculated is$$
\frac{\partial(s,t)}{\partial(x,y)}$$which is not the Jacobian you need, but luckily it is the reciprocal of what you need.

So get the x and y's out of that final integral and get the correct limits on it for s and t and you will be there.

 

[haruspex]

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

When changing variables, it's important to keep track of what variable the range refers to. The above reads like \int _{s=1}^4 \int _{t=x^2}^{2x^2} \frac {1}{t}\,ds \ dt, but that's wrong. What you have arrived at is:
\int _{s=1}^4 \int _{y=x^2}^{2x^2} \frac {1}{t}\,ds \ dt
Now substitute for y using s and t.

 

[Emspak]

The limits of integration are incorrect.

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

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

I need to get the x and y variables out of there. Since we said

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

so
y= \sqrt{s - x^2} and y = \frac{t}{x^2}

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

although i think i might have the order of integration wrong.

 

[haruspex]

t = \frac{y}{x^2} so y = \frac{t}{x^2}

No, try that again.

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

You're still not substituting the range correctly. If one end of the range is y = x², and you are substituting y = tx², what does that give you for t?

 

[Emspak]

so y = tx^{2 (that was bad algebra on my part) so the end of the range is t = y / x2, which would be the lower limit. SInce the upper limit is y = 2x2 and in terms of t that's 2x2 / y , yes?}

 

[SammyS]

so y = tx^{2 (that was bad algebra on my part) so the end of the range is t = y / x2, which would be the lower limit. SInce the upper limit is y = 2x2 and in terms of t that's 2x2 / y , yes?}

<sup>Look:

One of the boundaries of B is y = 2x2 .

But $\displaystyle \ t =\frac{y}{x^2} .$

Plug y = 2x2 into that.So t = ?</sup>

 

[Emspak]

Ah, so t = 2, right? Gawd. I feel like I am simply stupid after this.

 

[SammyS]

Ah, so t = 2, right? Gawd. I feel like I am simply stupid after this.

Hey. This is how we learn this stuff.


Now go back and review this thread -- in particular where LCKurtz came up with the change of coordinates.

While that set of coordinates follows pretty naturally for the region, B, notice that the expression you posted initially, before correcting the problem, -- that first expression still would have been a mess to work with, even with this nice set of coordinates.