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Iterated integral 2

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
\(\displaystyle \int_0^1\int_0^1 \frac{xy}{\sqrt{x^2+y^2+1}} dxdy\)
I start with subsitate \(\displaystyle u=x <=> du=dx\) and \(\displaystyle du= \frac{y}{\sqrt{x^2+y^2+1}} <=>u=y\ln\sqrt{x^2+y^2+1}\) so we got integrate by part that
\(\displaystyle xy\ln\sqrt{x^2+y^2+1}]_0^1-\int_0^1\frac{y}{\sqrt{x^2+y^2+1}}dx\)
and we got
\(\displaystyle [xy\ln\sqrt{x^2+y^2+1}]_0^1-[y\ln{\sqrt{x^2+y^2+1}}]_0^1\)
Remember that we solve dx. Is this correct?

Regards,
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Petrus,

I think I would first write the integral as:

\(\displaystyle \int_0^1 y\left(\int_0^1\frac{x}{\sqrt{x^2+y^2+1}}\,dx \right)\,dy\)

Now, on the inner integral, consider the substitution:

\(\displaystyle u=x^2+y^2+1\)

What do you have now?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hello MHB,
\(\displaystyle \int_0^1\int_0^1 \frac{xy}{\sqrt{x^2+y^2+1}} dxdy\)
I start with subsitate \(\displaystyle u=x <=> du=dx\) and \(\displaystyle du= \frac{y}{\sqrt{x^2+y^2+1}} <=>u=y\ln\sqrt{x^2+y^2+1}\) so we got integrate by part that
\(\displaystyle xy\ln\sqrt{x^2+y^2+1}]_0^1-\int_0^1\frac{y}{\sqrt{x^2+y^2+1}}dx\)
and we got
\(\displaystyle [xy\ln\sqrt{x^2+y^2+1}]_0^1-[y\ln{\sqrt{x^2+y^2+1}}]_0^1\)
Remember that we solve dx. Is this correct?

Regards,
Hi Petrus, :)

Substituting \(x=u\) won't give you anything useful since you are just replacing \(x\) by \(u\). You can try solving this problem with the substitution Mark has given, but let me suggest a slightly different method.

Differentiate \(\sqrt{x^2+y^2+1}\) and see what you get and try to use that result in solving the integral.
 

Petrus

Well-known member
Feb 21, 2013
739
Hello Petrus,

I think I would first write the integral as:

\(\displaystyle \int_0^1 y\left(\int_0^1\frac{x}{\sqrt{x^2+y^2+1}}\,dx \right)\,dy\)

Now, on the inner integral, consider the substitution:

\(\displaystyle u=x^2+y^2+1\)

What do you have now?
Hmm... \(\displaystyle u=x^2+y^2+1\) \(\displaystyle du =2x\)
\(\displaystyle \int_0^1 \frac{y}{2} \int_0^1 \frac{1}{\sqrt{u}}du\)
hmmm is this correct?

Regards,
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
Hi Petrus, :)

Substituting \(x=u\) won't give you anything useful since you are just replacing \(x\) by \(u\). You can try solving this problem with the substitution Mark has given, but let me suggest a slightly different method.

Differentiate \(\sqrt{x^2+y^2+1}\) and see what you get and try to use that result in solving the integral.
Hello Sudharaka,
I can't see how you can see this will give no progress. Have I miss something with 'integrate by part', I honestly doubt what method I will use when I come to this kind of question. Function divide by function or function multiplicate by function ( When you integrate)

Regards,
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Be careful to the boundaries of integration when you make a substitution.
 

Petrus

Well-known member
Feb 21, 2013
739
Be careful to the boundaries of integration when you make a substitution.
Yeah I am aware of that :) I will just subsitute back after I antiderivate, that's why I did not rewrite the limit of integrate

Regards,
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Hi Petrus, :)

Substituting \(x=u\) won't give you anything useful since you are just replacing \(x\) by \(u\).


I think what Petrus is trying to do is integration by parts .

If you differentiate \(\displaystyle y \sqrt {x^2+y^2+1} \) w.r.t to $ x$ what do we get ?
 

Petrus

Well-known member
Feb 21, 2013
739
So far I got:
\(\displaystyle u=x^2+y^2+1\) \(\displaystyle du =2x\)
\(\displaystyle \int_0^1 \frac{y}{2} \int_0^1 \frac{1}{\sqrt{u}}du\)
\(\displaystyle \int_0^1 \frac{y}{2} [2\sqrt{u}]\) If we subsitute it back we get:
\(\displaystyle \int_0^1 \frac{y}{2} [2\sqrt{x^2+y^2+1}]_0^1\)
now we got:
\(\displaystyle \int_0^1y\sqrt{2+y^2}-y\sqrt{1+y^2}\)
We subsitute \(\displaystyle u=2+y^2\) \(\displaystyle du=2y\)
so we got
\(\displaystyle \frac{1}{2}\int_0^1 \sqrt{u}-\sqrt{u-1}\)
If I antiderivate that I get
\(\displaystyle \frac{1}{2}[\frac{2u^{1.5}}{3}-\frac{2(u-1)^{1.5}}{3}]\)
and when i subsitute back and put the limits I get wrong answer, have I done something wrong here?

Regards,
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hello Sudharaka,
I can't see how you can see this will give no progress. Have I miss something with 'integrate by part', I honestly doubt what method I will use when I come to this kind of question. Function divide by function or function multiplicate by function ( When you integrate)

Regards,
Ah, you are trying to integrate by parts, that makes sense. I think I misunderstood your attempt to integrate by parts when reading the line,


........I start with subsitate \(\displaystyle u=x <=> du=dx\) and......
in your first post. Sorry about that. What I was suggesting is to observe that,

\[\frac{d}{dx}\sqrt{x^2+a}=\frac{x}{\sqrt{x^2+a}}\]

where \(a\) is a constant. That is,

\[\int \frac{x}{\sqrt{x^2+a}}\,dx=\sqrt{x^2+a}+C\]

where \(C\) is an arbitrary constant. I thought that this is something which isn't hard to observe. Using this you can solve the inner integral of your iterated integral.
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks Sudharaka,
I did never think about that! Can someone control my post #9 what I have done wrong?

Regards,
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Thanks Sudharaka,
I did never think about that! Can someone control my post #9 what I have done wrong?

Regards,
I dont see any mistakes .
 

Petrus

Well-known member
Feb 21, 2013
739
I dont see any mistakes .
I looked at wrong answer in facit haha.... Thanks got it now ;) I will try solve it with integrate by part now :)