MHB MHBSolving a Double Integral Problem with Polar Form

[Petrus]

Hello MHB,
I got as homework to solve this problem and get recommend to solve it with polar but I have not really work with polar but we have had lecture about it and I have done some research. This is the problem and what I understand
$$\int\int_Dx^3y^2\ln(x^2+y^2)$$, $$4\leq x^2+y^2\leq 25$$ and $$x,y\geq 0$$
if we want to change it to polar form let's write $$x=r\cos\theta$$ and $$y=r\sin\theta$$
so we got:
$$\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}r^3\cos^3\theta*r^2\sin^2\theta \ln(r^2\cos^2\theta+r^2\sin^2\theta) \ drd\theta$$
and we got our identity that $$x^2+y^2=r^2$$ that means we got our r limit as $$2\leq r \leq 5$$ if I am thinking correct we can't use our negative limit cause it says $$x,y\geq 0$$ I am stuck with how to get my $$\theta$$ limit well so far I can think we know that $$4 \leq r^2cos^2\theta + r^2sin^2\theta \leq 25$$ here is what I strugle with. is solve limit for $$\theta$$

Regards,

 

[I like Serena]

Hello MHB,
I got as homework to solve this problem and get recommend to solve it with polar but I have not really work with polar but we have had lecture about it and I have done some research. This is the problem and what I understand
$$\int\int_Dx^3y^2\ln(x^2+y^2)$$, $$4\leq x^2+y^2\leq 25$$ and $$x,y\geq 0$$
if we want to change it to polar form let's write $$x=r\cos\theta$$ and $$y=r\sin\theta$$
so we got:
$$\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}r^3\cos^3\theta*r^2\sin^2\theta \ln(r^2\cos^2\theta+r^2\sin^2\theta) \ drd\theta$$
and we got our identity that $$x^2+y^2=r^2$$ that means we got our r limit as $$2\leq r \leq 5$$ if I am thinking correct we can't use our negative limit cause it says $$x,y\geq 0$$ I am stuck with how to get my $$\theta$$ limit well so far I can think we know that $$4 \leq r^2cos^2\theta + r^2sin^2\theta \leq 25$$ here is what I strugle with. is solve limit for $$\theta$$

If x and y are both at least zero, that means each point is in the first quadrant.
In the first quadrant the bounds for $\theta$ are $0$ and $\dfrac \pi 2$.Furthermore, you should use the trigonometric identity $\cos^2\theta + \sin^2\theta=1$.In your original integral you omitted $dxdy$, which represent a very small square with sides of length $dx$ and $dy$.
In polar coordinates we also have a very small square-like region with sides $dr$ and $rd\theta$. The area of that small square is (in the limit) $rd\theta dr$.
You're supposed to replace $dxdy$ by $rd\theta dr$ when switching to polar coordinates, introducing an extra $r$.
This extra factor $r$ is called the Jacobian.

 

[Petrus]

If x and y are both at least zero, that means each point is in the first quadrant.
In the first quadrant the bounds for $\theta$ are $0$ and $\dfrac \pi 2$.Furthermore, you should use the trigonometric identity $\cos^2\theta + \sin^2\theta=1$.In your original integral you omitted $dxdy$, which represent a very small square with sides of length $dx$ and $dy$.
In polar coordinates we also have a very small square-like region with sides $dr$ and $rd\theta$. The area of that small square is (in the limit) $rd\theta dr$.
You're supposed to replace $dxdy$ by $rd\theta dr$ when switching to polar coordinates, introducing an extra $r$.
This extra factor $r$ is called the Jacobian.

Hello I like Serena
I understand they used that method when I saw it I was thinking and figoure it out, $\cos^2\theta + \sin^2\theta=1$
hmm this got more tricky then I thought when you say "In the first quadrant the bounds for $\theta$ are $0$ and $\dfrac \pi 2$." does that means that is our lower limit if I understand this correct? I need to think more about this

Regards,

 

[I like Serena]

Hello I like Serena
I understand they used that method when I saw it I was thinking and figoure it out, $\cos^2\theta + \sin^2\theta=1$
hmm this got more tricky then I thought when you say "In the first quadrant the bounds for $\theta$ are $0$ and $\dfrac \pi 2$." does that means that is our lower limit if I understand this correct? I need to think more about this

Regards,

Check out the picture I drew in post http://www.mathhelpboards.com/f10/integral-polar-coordinates-4389/index2.html#post19934 (I know you already did ;)).
The lower limit for $\theta$ is $0$.

 

[Petrus]

Check out the picture I drew in post http://www.mathhelpboards.com/f10/integral-polar-coordinates-4389/index2.html#post19934 (I know you already did ;)).
The lower limit for $\theta$ is $0$.

Ohh now I see, Thanks!

I know you already did ;))

I did not know I had a stalker

 

[Petrus]

Hello MHB,
I wanted to share the solution if someone is intressted ( did take me around 4h to integrate but I learned alot!)

problem:
calculate double integrate

,
there

We convert to polar cordinate (read post over how I did) and get:

$$\int_0^{\frac{\pi}{2}}\int_2^5 r^4\cos^3( \theta)r^2\sin( \theta) \ln(r^2\cos^2( \theta) +r^2\sin^2( \theta)) \ drd\theta$$

still look complicated for me and I want to try simplify as much as possible. We can use Fubini's theorem and take out our constant and we can use pythagorean trigonometric identity and rewrite that inside ln $$\ln(r^2\cos^2( \theta) +r^2\sin^2( \theta)) <=> \ln(r^2(\cos^2( \theta) +\sin^2( \theta))) <=> \ln(r^2(1))$$
so how does our integrate looks like now after we done all this!
$$\int_0^{\frac{\pi}{2}} \cos^3(\theta)sin^2(\theta) \ d\theta \int_2^5 r^6 \ln(r^2) \ dr$$
That looks a lot more better! let's start integrate our

$$d\theta$$:
We got
$$\int_0^{\frac{\pi}{2}} \cos^3(\theta)sin^2 (\theta) \ d\theta$$
Let's start to break it up so we got:
$$\int_0^{\frac{\pi}{2}} sin^2 (\theta) \cos^2(\theta)\cos( \theta) \ d\theta$$
Now we can use the identity $$\cos^2 ( \theta) =1- \sin^2( \theta)$$
and we got:
$$\int_0^{\frac{\pi}{2}} \sin^2(\theta)(1-\sin^2( \theta)) \cos (\theta) \ d\theta$$
let's simplify to:
$$\int_0^{\frac{\pi}{2}} (\sin^2(\theta)-\sin^4( \theta)) \cos (\theta) \ d\theta$$
now we can integrate! Subsitute $$u= \sin(\theta) <=> du= \cos(\theta)$$ (notice that I don't rewrite the limit cause I will subsitute back after I antidifferentiat) so we got
$$\int_0^{\frac{\pi}{2}} (u^2-u^4) du$$
if we antidifferentiat that and subsitute back we get
$$ \left[\frac{\sin^3(\theta)}{3}- \frac{\sin^5(\theta)}{5} \right]_0^{\frac{\pi}{2}} = \frac{2}{15}$$

dr:
$$\int_2^5 r^6 \ln(r^2) dr$$
use integrate by part and choose that
$$u=\ln(r^2) <=> du=\frac{2}{r} dr$$ and $$dv=r^6 dr <=> u= \frac{r^7}{7}$$
I will skip integrate by part and you will get result (without simplify)
$$\ln(25) \frac{5^7}{7} - \ln(4) \frac{2^7}{7} - \frac{1555994}{49}$$
now we have to multiplicate both of them so the result is
$$\frac{2}{15}(\ln(25) \frac{5^7}{7} - \ln(4) \frac{2^7}{7} - \frac{1555994}{49})$$

Regards,

 

[I like Serena]

Looks good! :)

 

[Petrus]

Looks good! :)

Thanks for helping me! The integration part did take me a lot atemp but it's start to become more easy, after practice and practice you start to get used with it :)

Regards,