# Homework Help: Double integration problem

1. Sep 10, 2016

1. The problem statement, all variables and given/known data
in this problem , i couldnt express all the x any y in terms of u( refer to the circled part ) ... so , i have problem to proceed my subsequent steps ....

2. Relevant equations

3. The attempt at a solution
i let u = (x^2) + (y^2) ...

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2. Sep 10, 2016

### BvU

1. Is this the full problem statement ?
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.

3. Sep 10, 2016

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

4. Sep 10, 2016

### BvU

What is the problem statement ?

5. Sep 10, 2016

This

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6. Sep 10, 2016

### BvU

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

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

7. Sep 10, 2016

### BvU

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

8. Sep 10, 2016

### Ray Vickson

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.

9. Sep 10, 2016

### 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

10. Sep 10, 2016

### Ray Vickson

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

11. Sep 11, 2016

or can I integrate in dx dy instead of dydx ?

12. Sep 11, 2016

### Staff: Mentor

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.

13. Sep 11, 2016

yes , i know the shape the graph , but it looks like i cant interchange the order of integration... I cant express x in terms of y ..
Well , i keep the dy/dx.... Back to the working in post # 1, i'm stucked at couldnt 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 ?

14. Sep 11, 2016

### Staff: Mentor

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.

15. Sep 11, 2016

i know and understand the shape of the region of integration...
But i still cant 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

16. Sep 11, 2016

### BvU

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 ?

17. Sep 11, 2016

Here

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18. Sep 11, 2016

### 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.

19. Sep 11, 2016

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

20. Sep 11, 2016

### BvU

There is an easy solution, don't worry.

21. Sep 11, 2016

yes
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 ???

22. Sep 11, 2016

### BvU

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 $x\over \sqrt{x^2 + y^2}$ ?

23. Sep 11, 2016

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

24. Sep 11, 2016

### BvU

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

25. Sep 12, 2016

I really hv no idea what it represent.....