Forum — Tournament Committee

In reply to Dirty_Player

PReDS

6/17/2010 1:01 PM UTC

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Hermit

You can also use the integral from 0 to b minus a, integral from 0 to b minus a dxdy. ...

No one is arguing – it's just a specific case of what I wrote.

Dirty_Player

I think the whole point is that the sides of the square are not parallel to the OX and OY axes.

Actually, the question from Rapper doesn't ask how to find the area of a figure bounded by lines with given equations. The question is how to find the area of a square using a multiple integral, so nothing prevents us from setting the coordinate axes parallel to its sides, and we can even place one of the vertices at the origin so that the figure is in the first quadrant, then the integrals will be from zero.
And even in the case of sides that are not parallel to the coordinate axes, an orthogonal transformation of the coordinate system (rotation) does not change the differential of the area, and with its help, the axes can be made parallel to the sides.

In reply to PReDS

Dirty_Player

6/17/2010 1:31 PM UTC

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PReDS

Actually, the question about the Rapper doesn't ask how to find the area of a figure bounded by lines with given equations. The question is how to find the area of a square using a multiple integral, so nothing prevents us from setting the coordinate axes parallel to its sides, and we can also place one of the vertices at the origin so that the figure is in the first quadrant, in which case the integrals will be from zero.

Well, it depends more on the conditions. If it's simple, then yes.

PReDS

And even in the case of sides not parallel to the coordinate axes, an orthogonal transformation of the coordinate system (rotation) does not change the differential of area, and with its help, the axes can be made parallel to the sides.

Coordinate transformation is a great idea. Although you can also play around with the equations of the lines :rolleyes:

Wicc

6/17/2010 1:37 PM UTC

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For those who enjoy double integrals, here's a problem: Compare which is greater: 1) the integral from [0;1] of x^x 2) the double integral over [0;1] * [0;1] of (xy)^(xy). The integral here is double, over a square. P.S. a^b means a to the power of b.

In reply to Dirty_Player

Hermit

6/17/2010 1:39 PM UTC

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Dirty_Player

It depends more on the conditions. If it's just a simple case, then yes.

Coordinate transformation is a great idea. Although you could also play around with the equations of lines :rolleyes:

In any case, you wouldn't need to know only the side; you would also need the angle of inclination.

Всё не так плохо как Вы думаете. Всё намного хуже!

In reply to Hermit

Dirty_Player

6/17/2010 1:54 PM UTC

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Hermit

In any case, it's not enough to know only the side; you also need the angle of inclination.

If we know the canonical equation of a line, then we have the tangent of the angle of inclination.

Wicc

Compare which is greater:
1) the integral from [0;1] of x^x
2) the integral over [0;1]*[0;1] of (xy)^(xy). The integral here is double, over a square.

Maybe I'm missing something, but in the first case, the integral is single, and in the second, it's double. It seems incorrect to compare length and area. Or did I misunderstand the condition?

In reply to Dirty_Player

Лось

6/17/2010 3:14 PM UTC

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"Maybe I'm missing something, but in the first case, the integral is single, and in the second, it's double. It seems incorrect to compare length and area. Or am I misunderstanding the problem?" A definite integral is a number. What do area and length have to do with it? Added after 3 minutes And the problem is simple.

In reply to Dirty_Player

Wicc

6/17/2010 3:57 PM UTC

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Dirty_Player

Maybe I'm missing something, but in the first case, the integral is single, and in the second, it's double. It seems incorrect to compare length and area. Or am I misunderstanding the problem?

There are no dimensions here, and both are just numbers. The idea is to compare these two numbers.

Added after 54 seconds

Лось

The problem is simple.

What's the answer?

Лось

6/17/2010 6:51 PM UTC

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In short, I got confused twice, so I'm sending the solution to Vika in a private message.

In reply to Лось

Dirty_Player

6/17/2010 6:58 PM UTC

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You need to enable logic. Vicc wouldn't have given this task if something was greater. Therefore, they are equal. The number is 0.783.

In reply to Dirty_Player

PReDS

6/17/2010 8:31 PM UTC

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Dirty_Player

You need to use logic. Wick wouldn't have given this task if something was greater. Therefore, they are equal. The number is 0.783 :)

Nonsense. I haven't dealt with this for ages, but in this situation, if my memory serves me right, the second integral can be represented as the product of 2 integrals: from the square of x and the square of y, i.e., it is essentially equal to the square of the first, the value of which is less than one. And 7th-grade math reminds us that a^2 < a, if 0 < a < 1.

P.S. And the numerical value of the first one is also incorrectly stated; it's only 1/3. Why are you feeding Rapper false information; he'll fail the exam.

Hermit

6/18/2010 5:57 AM UTC

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1/3, I think, is the value for *x²*, but for *x^X* it will be something else...

Всё не так плохо как Вы думаете. Всё намного хуже!

In reply to Hermit

PReDS

6/18/2010 7:33 AM UTC

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Hermit

I think 1/3 is the value for x², but for x^x it will be something else...

Sorry, I misread the condition due to poor eyesight.

In reply to PReDS

scorp

6/19/2010 6:58 AM UTC

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Maybe someone can help solve this riddle :D

How did the blue hero end up on the snow? There is no Dimension Door, no flight, and no wings; he wasn't freed from prison by a hero with wings (dd+flight) and then dismissed ... ?

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RomarioGrom

6/19/2010 8:31 AM UTC

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scorp

maybe someone can suggest a solution to the riddle

1 Superman :D
2 the second hero helped him get over the cliff :confused:
3 there was a hero with flight (book, scroll) who, after casting (on the mountain-scout), took (the scroll, the book) and ran away, and this scout flew over to a dead end ;)