Welcome to our community

Be a part of something great, join today!

Double integrals

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I got as homework to solve this problem and get recomend 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
\(\displaystyle \int\int_Dx^3y^2\ln(x^2+y^2)\), \(\displaystyle 4\leq x^2+y^2\leq 25\) and \(\displaystyle x,y\geq 0\)
if we want to change it to polar form lets write \(\displaystyle x=r\cos\theta\) and \(\displaystyle y=r\sin\theta\)
so we got:
\(\displaystyle \int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}r^3\cos^3\theta*r^2\sin^2\theta \ln(r^2\cos^2\theta+r^2\sin^2\theta) \ drd\theta\)
and we got our identity that \(\displaystyle x^2+y^2=r^2\) that means we got our r limit as \(\displaystyle 2\leq r \leq 5\) if I am thinking correct we can't use our negative limit cause it says \(\displaystyle x,y\geq 0\) I am stuck with how to get my \(\displaystyle \theta\) limit well so far I can think we know that \(\displaystyle 4 \leq r^2cos^2\theta + r^2sin^2\theta \leq 25\) here is what I strugle with. is solve limit for \(\displaystyle \theta\)

Regards,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Hello MHB,
I got as homework to solve this problem and get recomend 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
\(\displaystyle \int\int_Dx^3y^2\ln(x^2+y^2)\), \(\displaystyle 4\leq x^2+y^2\leq 25\) and \(\displaystyle x,y\geq 0\)
if we want to change it to polar form lets write \(\displaystyle x=r\cos\theta\) and \(\displaystyle y=r\sin\theta\)
so we got:
\(\displaystyle \int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}r^3\cos^3\theta*r^2\sin^2\theta \ln(r^2\cos^2\theta+r^2\sin^2\theta) \ drd\theta\)
and we got our identity that \(\displaystyle x^2+y^2=r^2\) that means we got our r limit as \(\displaystyle 2\leq r \leq 5\) if I am thinking correct we can't use our negative limit cause it says \(\displaystyle x,y\geq 0\) I am stuck with how to get my \(\displaystyle \theta\) limit well so far I can think we know that \(\displaystyle 4 \leq r^2cos^2\theta + r^2sin^2\theta \leq 25\) here is what I strugle with. is solve limit for \(\displaystyle \theta\)
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

Well-known member
Feb 21, 2013
739
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,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
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 #11 in this thread (I know you already did ;)).
The lower limit for $\theta$ is $0$.
 

Petrus

Well-known member
Feb 21, 2013
739

Petrus

Well-known member
Feb 21, 2013
739
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:

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

still look complicated for me and I wanna 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 \(\displaystyle \ln(r^2\cos^2( \theta) +r^2\sin^2( \theta)) <=> \ln(r^2(\cos^2( \theta) +\sin^2( \theta))) <=> \ln(r^2(1))\)
so how does our integrate looks like now after we done all this!
\(\displaystyle \int_0^{\frac{\pi}{2}} \cos^3(\theta)sin^2(\theta) \ d\theta \int_2^5 r^6 \ln(r^2) \ dr\)
That looks alot more better! let's start integrate our

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

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

Regards,

 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Looks good! :)
 

Petrus

Well-known member
Feb 21, 2013
739
Looks good! :)
Thanks for helping me! The integration part did take me alot atemp but it's start to become more easy, after practice and practice you start to get used with it :)

Regards,