Double Integral Solution for Iterated Integral 2

[Petrus]

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

Regards,

 

[MarkFL]

Hello Petrus,

I think I would first write the integral as:

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

Now, on the inner integral, consider the substitution:

$$u=x^2+y^2+1$$

What do you have now?

 

[Sudharaka]

Hello MHB,
$$\int_0^1\int_0^1 \frac{xy}{\sqrt{x^2+y^2+1}} dxdy$$
I start with subsitate $$u=x <=> du=dx$$ and $$du= \frac{y}{\sqrt{x^2+y^2+1}} <=>u=y\ln\sqrt{x^2+y^2+1}$$ so we got integrate by part that
$$xy\ln\sqrt{x^2+y^2+1}]_0^1-\int_0^1\frac{y}{\sqrt{x^2+y^2+1}}dx$$
and we got
$$[xy\ln\sqrt{x^2+y^2+1}]_0^1-[y\ln{\sqrt{x^2+y^2+1}}]_0^1$$
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]

Hello Petrus,

I think I would first write the integral as:

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

Now, on the inner integral, consider the substitution:

$$u=x^2+y^2+1$$

What do you have now?

Hmm... $$u=x^2+y^2+1$$ $$du =2x$$
$$\int_0^1 \frac{y}{2} \int_0^1 \frac{1}{\sqrt{u}}du$$
hmmm is this correct?

Regards,

 

[Petrus]

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,

 

[alyafey22]

Be careful to the boundaries of integration when you make a substitution.

 

[Petrus]

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,

 

[alyafey22]

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 $$ y \sqrt {x^2+y^2+1} $$ w.r.t to $ x$ what do we get ?

 

[Petrus]

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

Regards,

 

[Sudharaka]

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 $$u=x <=> du=dx$$ 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]

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

Regards,

 

[alyafey22]

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 .

 

[Petrus]

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 :)