Evaluating Double Integral for Region Ω

[DryRun]

Homework Statement
Evaluate $$\int\int (2x+1)(x-y)dxdy$$ where Ω is the region in the first quadrant of the x-y plane bounded by: y=x, y=x+2, y=2-x^2 and y=4-x^2

The attempt at a solution
This is the graph of the intersecting curves and the shaded area is the region Ω:
http://s2.ipicture.ru/uploads/20120324/RJSv46D4.png

Description of region Ω requires adding 2 parts of the region:

Left-hand side(region A):
For x fixed, y varies from y=2-x^2 to y=x+2
x varies from x=0 to x=1

Right-hand side (region B):
For x fixed, y varies from y=x to y=4-x^2
x varies from x=1 to x=1.56

Now, for the region A:
$$\int_0^1 \int_{2-x^2}^{x+2} (2x+1)(x-y)dydx$$

And for region B:
$$\int_1^{1.56} \int_x^{4-x^2} (2x+1)(x-y)dydx$$

Therefore,
$$\int\int (2x+1)(x-y)dxdy = \int_0^1 \int_{2-x^2}^{x+2} (2x+1)(x-y)dydx + \int_1^{1.56} \int_x^{4-x^2} (2x+1)(x-y)dydx$$

Evaluating double integral for region A:
$$\int_0^1 \int_{2-x^2}^{x+2} (2x+1)(x-y)dydx$$
$$= \int_0^1 \int_{2-x^2}^{x+2} (2x^2+x-2xy-y)dydx$$
$$\int (2x^2+x-2xy-y)dy=2x^2y+xy-xy^2-\frac{y^2}{2}$$
$$\int_{2-x^2}^{x+2} (2x^2+x-2xy-y)dy=2x^2(x+2)+x(x+2)-x(x+2)^2-\frac{(x+2)^2}{2}-[2x^2(2-x^2)+x(2-x^2)-x(2-x^2)^2-\frac{(2-x^2)^2}{2}]$$

This is the part where the calculations get too crowded. I'm wondering if there's a shortcut or better way of tackling the whole evaluation?

OK, i pushed on as i don't know any other methods:
$$\int_{2-x^2}^{x+2} (2x^2+x-2xy-y)dy=x^5+\frac{5x^4}{2}-2x^3-\frac{11x^2}{2}-2x$$
Now,
$$\int_0^1 (x^5+\frac{5x^4}{2}-2x^3-\frac{11x^2}{2}-2x)dx=-\frac{8}{3}$$

Evaluating double integral for region B:
$$\int_1^{1.56} \int_x^{4-x^2} (2x+1)(x-y)dydx$$
$$\int_x^{4-x^2} (2x+1)(x-y)dy=-x^5-\frac{5x^4}{2}+7x^3+\frac{21x^2}{2}-12x-8$$
Now,
$$\int_1^{1.56} (-x^5-\frac{5x^4}{2}+7x^3+\frac{21x^2}{2}-12x-8).dx$$
$$\int (-x^5-\frac{5x^4}{2}+7x^3+\frac{21x^2}{2}-12x-8).dx=-\frac{x^6}{6}-\frac{x^5}{2}+\frac{7x^4}{4}+\frac{7x^3}{2}-6x^2-8x$$
So,
$$\int_1^{1.56} (-x^5-\frac{5x^4}{2}+7x^3+\frac{21x^2}{2}-12x-8).dx=-10.45153638-(-\frac{113}{12})=-1.034869713$$

Therefore, the answer is:
$$\int\int (2x+1)(x-y)dxdy = -\frac{8}{3}-1.034869713=-3.70153638$$

 

[LCKurtz]

You have set up the problem perfectly. Personally I wouldn't waste my time actually working it out like that. That type of problem is perfect for something like Maple, which gives an answer of -4 for the sum of both integrals.

The only alternative that comes to mind is some clever change of variables to map the region to a rectangle or something like that. Did this question arise in a problem section about change of variables in multiple integrals?

 

[DryRun]

Hi LCKurtz

Actually, it's a question from my maths test that i wasn't able to complete due to time restrictions. The problem expanded so much on my answer sheet that it drove me to desperation, so i was wondering that maybe my examiner had something else in mind (some other simpler method) that i missed. And there was no mention of change of variables in the question. It was a standalone question.

 

[LCKurtz]

Hi LCKurtz

Actually, it's a question from my maths test that i wasn't able to complete due to time restrictions. The problem expanded so much on my answer sheet that it drove me to desperation, so i was wondering that maybe my examiner had something else in mind (some other simpler method) that i missed. And there was no mention of change of variables in the question. It was a standalone question.

That is a ridiculous question to appear in a timed exam. A take-home, maybe. Are you sure the question didn't say to just set up the integrals with correct limits? Then it would be appropriate, and you had it correctly done right off the bat.

 

[DryRun]

The question stated: evaluate the double integral. Maybe my examiner hates my class.

The final answer that i got is −3.70153638 and your answer says -4. Maybe it's because i approximated the limit of region B to 1.56?

The exact x-coordinate is: $$\frac{-1+\sqrt{17}}{2}=1.561552813$$
If i had evaluated this integral instead:
$$\int_1^{1.561552813} \int_x^{4-x^2} (2x+1)(x-y)dydx$$
Maybe i would have got the exact final answer: -4.

 

[LCKurtz]

The final answer that i got is −3.70153638 and your answer says -4. Maybe it's because i approximated the limit of region B to 1.56?
The exact x-coordinate is: $$\frac{-1+\sqrt{17}}{2}=1.561552813$$

I used that exact value in Maple. Maple actually gave -4.00000001 and I'm guessing the exact answer is -4.

You obviously know how to set it up correctly and work integrals. So unless it really bothers you, I wouldn't waste any more time on it. If your test key comes back with a solution involving some clever change of variables, I would be interested to see the solution.

 

[DryRun]

I'll respond to this thread with feedback as soon as the test answers are posted on my faculty's website.

Thank you for your help, LCKurtz.

 

[DryRun]

OK, the answer for the test question in post #1 has been released, and the solution (surprisingly short!) involved making use of transformation and mapping.
$$Let \,u=y-x \\Let \,v=y+x^2$$
From there, the Jacobian is found, and its reciprocal gives:
$$dxdy=\frac{1}{1+2x}\,.dudv$$
The mapping on u-v axes, gives a square graph, where u varies from u=0 to u=2 and v varies from v=2 to v=4.
The final answer is exactly -4.

I have no idea how the transformations for u and v were derived based on the 4 equations in x-y (that's something completely new to me as i didn't even know that this was somehow possible!?), although i can vaguely make it out from the original equations, but it still doesn't make much sense to me. There should be some solid method to easily derive and find the correct transformation, but it eludes me. I have to say that i have never encountered a problem like that in my notes or the many problems that I've worked through before. I believe that the transformation equations should have been given as part of the problem, but maybe they were omitted by mistake? I can't gather enough courage to go knock at my lecturer's office and suggest that he might have overlooked those u-v equations, so i guess i'll just let this one hang as disappointing history, unless someone can come up with a satisfying explanation on how those u-v equations were derived?

 

[LCKurtz]

OK, the answer for the test question in post #1 has been released, and the solution (surprisingly short!) involved making use of transformation and mapping.
$$Let \,u=y-x \\Let \,v=y+x^2$$
From there, the Jacobian is found, and its reciprocal gives:
$$dxdy=\frac{1}{1+2x}\,.dudv$$
The mapping on u-v axes, gives a square graph, where u varies from u=0 to u=2 and v varies from v=2 to v=4.
The final answer is exactly -4.

I have no idea how the transformations for u and v were derived based on the 4 equations in x-y (that's something completely new to me as i didn't even know that this was somehow possible!?), although i can vaguely make it out from the original equations, but it still doesn't make much sense to me. There should be some solid method to easily derive and find the correct transformation, but it eludes me. I have to say that i have never encountered a problem like that in my notes or the many problems that I've worked through before. I believe that the transformation equations should have been given as part of the problem, but maybe they were omitted by mistake? I can't gather enough courage to go knock at my lecturer's office and suggest that he might have overlooked those u-v equations, so i guess i'll just let this one hang as disappointing history, unless someone can come up with a satisfying explanation on how those u-v equations were derived?

Well, I suspected a change of variables was in order. I probably should have spotted that. If you write your boundary equations with the variables on one side, they become:
$y-x = 0,\ y-x = 2,\ y+x^2=2,\ y+x^2=4$ That certainly suggests the change of variables $u = y-x,\ v=y+x^2$ would map the area to a rectangle.

 

[DryRun]

So, u and v each have 2 distinct values? It's a bit confusing, as u and v should normally each have a single fixed value. For example, (y-x) should give a fixed value, but here we have two values; 0 and 2.
I'm thinking, maybe an alternative (and more comprehensible option) would be:
$Let\ u_1 = y-x \\Let\ u_2= y-x$

Are there any limitations to deriving the transformations using that method? It's a tricky problem as I've never had to derive the transformations, since they are always given in the problems I've encountered until now.

For example, let's consider:
$y-x=0,\ y-x=2x^2$
In that case, how would we map the transformations? Do we still use $Let\ u=y-x$?

 

[LCKurtz]

Well, I suspected a change of variables was in order. I probably should have spotted that. If you write your boundary equations with the variables on one side, they become:
$y-x = 0,\ y-x = 2,\ y+x^2=2,\ y+x^2=4$ That certainly suggests the change of variables $u = y-x,\ v=y+x^2$ would map the area to a rectangle.

So, u and v each have 2 distinct values? It's a bit confusing, as u and v should normally each have a single fixed value. For example, (y-x) should give a fixed value, but here we have two values; 0 and 2.
I'm thinking, maybe an alternative (and more comprehensible option) would be:
$Let\ u_1 = y-x \\Let\ u_2= y-x$

Are there any limitations to deriving the transformations using that method? It's a tricky problem as I've never had to derive the transformations, since they are always given in the problems I've encountered until now.

Your teacher "cooked" this problem up to make it work out. With the boundaries of your region given as $y-x = 0,\ y-x = 2,\ y+x^2=2,\ y+x^2=4$ the change of variables $u = y-x,\ v=y+x^2$ gives new $u,v$ boundaries $u=0,\ u=2,\ v=2,\ v=4$ which is a rectangle in the $uv$ plane. Now the Jacobian is$$
J=\frac{\partial(u,v)}{\partial(x,y)}=\left |\begin{array}{cc}-1&1\\-2x&1\\
\end{array}\right | = -1-2x$$ So using the absolute value you have the change of variables formula$$
dydx = \frac 1 {2x+1}dudv$$This is the point where typically you would have to solve your two equations for $x$ and $y$ in terms of $u$ and $v$ to perform the substitution which, in this problem, would require using the quadratic formula and making a mess of the problem. (This is why I didn't really examine the substitution method when I first looked at your problem). But the original integral was$$
\iint_R(2x+1)(x-y)dydx$$Look what happens when you substitute in for $dxdy$. You get$$
\int_2^4\int_0^2(x-y)dudv$$ because the $2x+1$ factors cancel. You would still normally have to substitute in the $x$ and $y$, but what luck, that $(x-y)$ just happens to be $-u$, so your new integral is$$
\int_2^4\int_0^2(-u)dudv$$So this problem works magically, without ever having to back-solve your substitution. That's why I say it is a "cooked up" example.

 

[DryRun]

I had tried to work it out again when i first looked at the solution, but i got stuck at substituting for x and y in (x-y) as it became too complicated. It didn't occur to me that (-u) did the trick!

Unfortunately, the "cook up" literally cooked up everyone's grades for that test... Hopefully, my end-of-year exams in 3 weeks won't be so tough.

Thank you, LCKurtz, for your clear explanation. Your help is much appreciated.