Solving a Double Integration Problem with Unknown Variables: How to Proceed?

[chetzread]

Homework Statement

in this problem , i couldn't express all the x any y in terms of u( refer to the circled part ) ... so , i have problem to proceed my subsequent steps ...

Homework Equations

The Attempt at a Solution

i let u = (x^2) + (y^2) ... [/B]

 

[BvU]

1. Is this the full problem statement ?
Use the template. I can hardly read your -2 lower bound.
Your u is a somewhat unfortunate choice. This whole thing reeks like something with a circle, so don't pick r^2 but pick r.

 

[chetzread]

1. Is this the full problem statement ?
Use the template. I can hardly read your -2 lower bound.
Your u is a somewhat unfortunate choice. This whole thing reeks like something with a circle, so don't pick r^2 but pick r.

what do you mean ? do you mean i pick the u wrongly ? Then , what should be the correct one ?

 

[BvU]

What is the problem statement ?

 

[chetzread]

What is the problem statement ?

This

 

[BvU]

That is not a problem statement. But I suppose you have to evaluate the integral ?

what do you mean ? do you mean i pick the u wrongly ? Then , what should be the correct one ?

I said unfortunate and you can read in post #2 what my suggestion for a better choice would be.

 

[BvU]

Let me give you another hint: draw a picture of the integration limits in the x-y plane

 

[Ray Vickson]

Homework Statement

in this problem , i couldn't express all the x any y in terms of u( refer to the circled part ) ... so , i have problem to proceed my subsequent steps ...

Homework Equations

The Attempt at a Solution

i let u = (x^2) + (y^2) ... [/B]

Just so you know: most helpers will not bother to look at posted images like you have supplied. If you genuinely want help you should take the trouble to actually type out your work. I suggest that you read the message "Guidelines for Students and Helpers" by Vela, which is pinned to the start of this Forum.

 

[BvU]

Ray is right (of course, he always is), but I grant you that typesetting something like $$
\int\limits_{-2}^{\ 2} \ \int\limits_0^{\ \ \sqrt{4-x^2}} {x\over\sqrt{x^2+y^2} } \ dy \;dx $$ isn't for the faint-hearted (*)

Now, how about my suggestion in post #7 ?That ring a bell ?

(*) Right-click the expression and pick Show math as $\ \ TeX$ commands

 

[Ray Vickson]

Ray is right (of course, he always is), but I grant you that typesetting something like $$
\int\limits_{-2}^{\ 2} \ \int\limits_0^{\ \ \sqrt{4-x^2}} {x\over\sqrt{x^2+y^2} } \ dy \;dx $$ isn't for the faint-hearted (*)

Now, how about my suggestion in post #7 ?That ring a bell ?

(*) Right-click the expression and pick Show math as $\ \ TeX$ commands

This can be done fairly easily also in plain text:
int_{x=-2..2} int_{y=0..sqrt(4-x^2)} x /sqrt(x^2+y^2) dy dx.
That is perfectly readable, although, of course, not as nice as LaTeX. One can even use the formula menu ribbon at the top of the input panel to write it as
∫_{-2..2} ∫_{0..√(4-x²)} x/√(x²+y²) dy dx

 

[chetzread]

or can I integrate in dx dy instead of dydx ?

 

[Mark44]

or can I integrate in dx dy instead of dydx ?

You can change the order of integration in an iterated integral. Of course, this means that the limits of integration will have to change. And this means that you have to understand the geometric region over which integration is to be performed.

 

[chetzread]

You can change the order of integration in an iterated integral. Of course, this means that the limits of integration will have to change. And this means that you have to understand the geometric region over which integration is to be performed.

yes , i know the shape the graph , but it looks like i can't interchange the order of integration... I can't express x in terms of y ..
Well , i keep the dy/dx... Back to the working in post # 1, I'm stucked at couldn't express all the x any y in terms of u( refer to the circled part ) ...
how to continue ? Or is there anything wrong in my working ?

 

[Mark44]

yes , i know the shape the graph

What graph? You need to understand the shape of the region of integration. I'm not talking about the integrand $\frac x {\sqrt{x^2 + y^2}}$, if that's what you meant.

, but it looks like i can't interchange the order of integration... I can't express x in terms of y ..

You might need to split the integral into two integrals.

Well , i keep the dy/dx... Back to the working in post # 1, I'm stucked at couldn't express all the x any y in terms of u( refer to the circled part ) ...
how to continue ? Or is there anything wrong in my working ?

Your work is a mess. On the next to last line your integrand involves x, y, and u.

I haven't worked the problem, but switching the order of integration definitely seems like the way to go.

 

[chetzread]

What graph? You need to understand the shape of the region of integration. I'm not talking about the integrand $\frac x {\sqrt{x^2 + y^2}}$, if that's what you meant.
You might need to split the integral into two integrals.

Your work is a mess. On the next to last line your integrand involves x, y, and u.

I haven't worked the problem, but switching the order of integration definitely seems like the way to go.

i know and understand the shape of the region of integration...
But i still can't proceed although trying many times...Can you show your working so that i can compare mine and yours ...thus , i can know which part i wrong

 

[BvU]

i know and understand the shape of the region of integration

Reassure us and post a picture (this time a photo is acceptable; we must compromise now and then ).

...thus , i can know which part i wrong

Showing the right way doesn't help in this. But I can tell you where you go wrong ( as was already mentioned:

On the next to last line your integrand involves x, y, and u

)

If you want to substitute an integration variable, in this case replace $dy$ by $du$, you need to

1. replace the bound of the variable to be substituted by the bound of the new variable: if $y$ runs from $0$ to $\ \sqrt {4-x^2}\$, then $u$ runs from $0$ to ...
2. eliminate $y$ completely from the expression, meaning you would end up with something like $$
   \int\limits_{-2}^{\ 2} \ \int\limits_0^{\ \ ...} {x\over\sqrt{u}} \ {1\over 2 \sqrt{u-x^2}} \ du \;dx
   $$

Which is equivalent to working from something that looks difficult to something that's nearly impossible.

Does your textbook in the preceding chapter have examples that bear some resemblance to the integral you are trying to evaluate in this thread ? In particular: an example or a discussion involving polar coordinates ?

 

[chetzread]

Reassure us and post a picture (this time a photo is acceptable; we must compromise now and then ).

Showing the right way doesn't help in this. But I can tell you where you go wrong ( as was already mentioned:

)

If you want to substitute an integration variable, in this case replace $dy$ by $du$, you need to

1. replace the bound of the variable to be substituted by the bound of the new variable: if $y$ runs from $0$ to $\ \sqrt {4-x^2}\$, then $u$ runs from $0$ to ...
2. eliminate $y$ completely from the expression, meaning you would end up with something like $$
   \int\limits_{-2}^{\ 2} \ \int\limits_0^{\ \ ...} {x\over\sqrt{u}} \ {1\over 2 \sqrt{u-x^2}} \ du \;dx
   $$

Which is equivalent to working from something that looks difficult to something that's nearly impossible.

Does your textbook in the preceding chapter have examples that bear some resemblance to the integral you are trying to evaluate in this thread ? In particular: an example or a discussion involving polar coordinates ?

Here

 

[BvU]

Thank you for the quick reply. That is step zero.
Next step there: draw a point $x,y$ and guess what the integrand represents in that picture ?

Re next paragraphs in post #16 :
Do you now understand the problems were with your next step ?
Do you feel convinced that your substitution $u$ does not make things easier ?

Did you bother to follow up on my suggestion in post #2 and the reminder in post #9 ?

There were more questions in the last paragraph in post #16.

 

[chetzread]

Reassure us and post a picture (this time a photo is acceptable; we must compromise now and then ).

Showing the right way doesn't help in this. But I can tell you where you go wrong ( as was already mentioned:

)

If you want to substitute an integration variable, in this case replace $dy$ by $du$, you need to

1. replace the bound of the variable to be substituted by the bound of the new variable: if $y$ runs from $0$ to $\ \sqrt {4-x^2}\$, then $u$ runs from $0$ to ...
2. eliminate $y$ completely from the expression, meaning you would end up with something like $$
   \int\limits_{-2}^{\ 2} \ \int\limits_0^{\ \ ...} {x\over\sqrt{u}} \ {1\over 2 \sqrt{u-x^2}} \ du \;dx
   $$

Which is equivalent to working from something that looks difficult to something that's nearly impossible.

Does your textbook in the preceding chapter have examples that bear some resemblance to the integral you are trying to evaluate in this thread ? In particular: an example or a discussion involving polar coordinates ?

No, so there's no solution for this problem?

 

[BvU]

There is an easy solution, don't worry.

 

[chetzread]

Do you feel convinced that your substitution uuu does not make things easier ?

yes

Does your textbook in the preceding chapter have examples that bear some resemblance to the integral you are trying to evaluate in this thread ? In particular: an example or a discussion involving polar coordinates ?

Next step there: draw a point x,yx,yx,y and guess what the integrand represents in that picture ?

what do you mean ? do you ask me to draw out
\int\limits_{-2}^{\ 2} \ \int\limits_0^{\ \ \sqrt{4-x^2}} {x\over\sqrt{x^2+y^2} } \ dy \;dx ?

 

[BvU]

what do you mean ? do you ask me to draw out
$$\int\limits_{-2}^{\ 2} \ \int\limits_0^{\ \ \sqrt{4-x^2}} {x\over\sqrt{x^2+y^2} } \ dy \;dx $$ ?

No/Yes. The integrand is not the integral. The integrand is $$x\over \sqrt{x^2 + y^2} $$ Take a point $x,y$; what is $x\over \sqrt{x^2 + y^2}$ ?

 

[chetzread]

No/Yes. The integrand is not the integral. The integrand is $$x\over \sqrt{x^2 + y^2} $$ Take a point $x,y$; what is $x\over \sqrt{x^2 + y^2}$ ?

the intergrand has nothing to do with the limit , right ?

 

[BvU]

Right. Did you draw a point $\ x,y \$ and consider what $\ x\over \sqrt{x^2 + y^2}$ represents ?

 

[chetzread]

Right. Did you draw a point $\ x,y \$ and consider what $\ x\over \sqrt{x^2 + y^2}$ represents ?

I really hv no idea what it represent...

 

[BvU]

Pity you don't post your drawing ...

 

[chetzread]

do u ask me to draw out $x\over \sqrt{x^2 + y^2}$ ?
If so , i really dun know how to draw it out ...

 

[chetzread]

Pity you don't post your drawing ...

this one ?

 

[BvU]

No, this one:

draw a point $\ x,y \$ and consider what $\ x\over \sqrt{x^2 + y^2}$ represents

Since words fail me, I will try it with a picture:

[edit] your picture looks cute, but why do you set the ranges to -50,50 instead of x -2,2 and y 0,2 ?

 

[chetzread]

No, this one:
Since words fail me, I will try it with a picture:
View attachment 105789

[edit] your picture looks cute, but wht do you set the ranges to -50,50 instead of x -2,2 and y 0,2 ?

So, how to proceed from ur diagram?

 

[BvU]

You answer the question

what $\ x\over \sqrt{x^2 + y^2}$represents

 

[chetzread]

You answer the question

it has no meaning , right ? the author just simply put inside the equation, right ?

 

[BvU]

You think I annoy you with so many questions when there is no meaning ? What is the sine of the red angle ? What is the cosine ?

 

[chetzread]

You think I annoy you with so many questions when there is no meaning ? What is the sine of the red angle ? What is the cosine ?
View attachment 105794

You answer the question

It's cos theta, so?

 

[Ray Vickson]

It's cos theta, so?

What do you think?