A doubt about double integration

[Theatre Of Fate]

Salutations!

This is my first post. I´m writing here because I have a doubt regarding a solved problem of double integration present in a book of mine. I don´t speak english yet ( ), but I will try to translate the problem to english. Well, here I go:

"Calculate int[ (x^2 + y^2) dx dy ] over the region D on the first quadrant of the xy-plane limitated by the hyperboles x^2 - y^2 = 1, x^2 - y^2 = 9, x*y = 2 and x*y = 4".

Well, using the following transformation: u = x^2 - y^2, v = x*y; we obtain the following Jacobian: J(u,v) = 1 / ( 2 * (u^2 + 4*v^2)^(1/2) ).

The region of integration Q on the uv-plane is the rectangle: Q = {(u,v) E R² | 1 <= u <= 9 , 2 <= v <= 4}.

So,

int[ (x^2 + y^2) dx dy ] = int[ (u^2 + 4*v^2) * ( 1 / ( 2 * (u^2 + 4*v^2)^(1/2) ) ) du dv ] = (1/2) * int[ du dv ] = 8.

Well, I don´t understand this solution. If you plot the region Q, you will obtain a rectangle lying on the uv-plane, but this rectangle includes (through the previous change of coordinates) the two regions lying on the xy-plane that are limitated by the four hyperboles: the first one lying on the first quadrant, and the second one lying on the third quadrant. But I want only the region on the first quadrant. I can´t see how the definition of the rectangle Q could exclude the region on the third quadrant. And this region must to be excluded, because the change of coordinates must to be injective (otherwise I can´t use the previous method to calculate double integrals).

I don´t know if you could understand what I wrote, or even if my doubt was clearly exposed (in case of my horrible english have been understood). But I thank in advance for any help!

 

[Wong]

Theatre of Fate, welcome to physics forums!

I don't speak english neither, so I wish you would understand what I say.

I think what I get from your post is that you are not at all sure that the transformation you use is one-one (because each hyperbola has two disconnected components and each point in (u,v) seems to "correspond" to two points in (x,y) ) and so you are not sure that the "change of variable" formula for integration works.

I think the point is that we may "restrict" the map/transformation to the first quadrant of the (x,y) plane. Then the map will be one-one and there will be no problem. Specifically, when calculating the Jacobian of the transformation, you need to express (x,y) in terms of (u,v). The transformation is, u = x^2 - y^2, v = x*y. When expressing (x,y) in (u,v), you will find that there is more than one solution (differring by +/- sign). We choose specifically the solution which maps (u,v) to the first quadrant of (x,y) plane and we will get the Jacobian as you mentioned.

Hope that helps.

 

[Ethereal]

Here's another question regarding double integration:

I was reading through a proof showing that the area under the standard normal curve = 1, and there's one part of the proof I couldn't follow:

$$I_{x} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$$

$$I_{y} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dy$$

Then you multiply $$I_{x}$$ with $$I_{y}$$:
$$I_{x} \cdot I_{y} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dxdy$$

Then the book goes on to say (or at least I think it says this since I don't have it with me now):

Since I(x) and I(y) are the same except for the choice of variables:

$$I^2 = \int_ {-\infty}^\infty \int_ {-\infty}^\infty \frac{1}{\sigma^2 2\pi}e^{\frac{-(x-\mu)^2-(y-\mu)^2}{2\sigma^2}$$

Why is the last step valid? I couldn't find a proof for it.

 

[rgoudie]

I don´t speak english yet

I don't speak english neither

Let me just say that you both speak English well enough to be easily understood.

-Ray.

 

[arildno]

Here's another question regarding double integration:

I was reading through a proof showing that the area under the standard normal curve = 1, and there's one part of the proof I couldn't follow:

$$I_{x} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$$

$$I_{y} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dy$$

Then you multiply $$I_{x}$$ with $$I_{y}$$:
$$I_{x} \cdot I_{y} = \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \int_ {-\infty}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dxdy$$

Then the book goes on to say (or at least I think it says this since I don't have it with me now):

Since I(x) and I(y) are the same except for the choice of variables:

$$I^2 = \int_ {-\infty}^\infty \int_ {-\infty}^\infty \frac{1}{\sigma^2 2\pi}e^{\frac{-(x-\mu)^2-(y-\mu)^2}{2\sigma^2}$$

Why is the last step valid? I couldn't find a proof for it.

Let's first look at a somewhat more general case:
The assertion is that (for constants a,b,c,d):
$$\int_{a}^{b}\int_{c}^{d}f(x)g(y)dxdy=\int_{a}^{b}f(x)(\int_{c}^{d}g(y)dy)dx=$$
$$\int_{c}^{d}g(y)(\int_{a}^{b}f(x)dx)dy=\int_{a}^{b}f(x)dx\int_{c}^{d}g(y)dy$$

The equality between the 1. and the 2.&3. terms are consequences of Fubini's theorem (essentially, that double integrals can be computed stepwise; that is we may first integrate with respect to one variable, keeping the other constant, and after that, integrating with respect to the last variable)
. The equality between the 2&3. terms and the 4.term is a consequence that the integral placed in parentheses within the other integral is a constant,
which therefore may be extracted from the outer integral.

 

[Ethereal]

Let's first look at a somewhat more general case:
The assertion is that (for constants a,b,c,d):
$$\int_{a}^{b}\int_{c}^{d}f(x)g(y)dxdy=\int_{a}^{b}f(x)(\int_{c}^{d}g(y)dy)dx=$$
$$\int_{c}^{d}g(y)(\int_{a}^{b}f(x)dx)dy=\int_{a}^{b}f(x)dx\int_{c}^{d}g(y)dy$$

The equality between the 1. and the 2.&3. terms are consequences of Fubini's theorem (essentially, that double integrals can be computed stepwise; that is we may first integrate with respect to one variable, keeping the other constant, and after that, integrating with respect to the last variable)
. The equality between the 2&3. terms and the 4.term is a consequence that the integral placed in parentheses within the other integral is a constant,
which therefore may be extracted from the outer integral.

Thanks for the reply, but could you explain this in more detail? I don't get what you are trying to say. Thanks.

 

[HallsofIvy]

Your English is excellent. Its my (put the language of your choice here) that is terrible!

 

[Theatre Of Fate]

Theatre of Fate, welcome to physics forums!

I don't speak english neither, so I wish you would understand what I say.

I think what I get from your post is that you are not at all sure that the transformation you use is one-one (because each hyperbola has two disconnected components and each point in (u,v) seems to "correspond" to two points in (x,y) ) and so you are not sure that the "change of variable" formula for integration works.

I think the point is that we may "restrict" the map/transformation to the first quadrant of the (x,y) plane. Then the map will be one-one and there will be no problem. Specifically, when calculating the Jacobian of the transformation, you need to express (x,y) in terms of (u,v). The transformation is, u = x^2 - y^2, v = x*y. When expressing (x,y) in (u,v), you will find that there is more than one solution (differring by +/- sign). We choose specifically the solution which maps (u,v) to the first quadrant of (x,y) plane and we will get the Jacobian as you mentioned.

Hope that helps.

Wong, I thank you for the explanation!