Double Integral Solution for Iterated Integral 2

  • MHB
  • Thread starter Petrus
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In summary, Petrus was trying to integrate y\ln\sqrt{x^2+y^2+1} by parts and substituted x=u. Differentiating y\sqrt{x^2+y^2+1}, he found that u=x^2+y^2+1 du=2x. He then applied the substitution u=2+y^2 du=2y to get \frac{1}{2}\int_0^1 \sqrt{u}-\sqrt{u-1}
  • #1
Petrus
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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,
 
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  • #2
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?
 
  • #3
Petrus said:
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.
 
  • #4
MarkFL said:
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,
 
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  • #5
Sudharaka said:
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,
 
  • #6
Be careful to the boundaries of integration when you make a substitution.
 
  • #7
ZaidAlyafey said:
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,
 
  • #8
Sudharaka said:
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 ?
 
  • #9
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:
  • #10
Petrus said:
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,

Petrus said:

...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.
 
  • #11
Thanks Sudharaka,
I did never think about that! Can someone control my post #9 what I have done wrong?

Regards,
 
  • #12
Petrus said:
Thanks Sudharaka,
I did never think about that! Can someone control my post #9 what I have done wrong?

Regards,

I don't see any mistakes .
 
  • #13
ZaidAlyafey said:
I don't 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 :)
 

FAQ: Double Integral Solution for Iterated Integral 2

What is an iterated integral?

An iterated integral is a type of integral that involves multiple variables and multiple integration steps. It is used to calculate the volume under a multi-variable function over a specific domain.

How is an iterated integral evaluated?

An iterated integral is evaluated by first integrating with respect to one variable, while holding the other variables constant. The resulting expression is then integrated with respect to another variable, and this process is repeated until all variables have been integrated.

What is the difference between a single integral and an iterated integral?

A single integral involves integrating a function with respect to one variable over a specific interval. An iterated integral involves integrating a multi-variable function with respect to multiple variables over a specific domain.

What is the purpose of using iterated integrals?

Iterated integrals are used to calculate the volume under a multi-variable function over a specific domain. They are also used in various applications in physics, engineering, and economics.

What are some common techniques used to solve iterated integrals?

Some common techniques used to solve iterated integrals include using the Fundamental Theorem of Calculus, changing the order of integration, and using substitution or integration by parts.

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