Jan 2011
14
0
Newcastle
I'm normally ok with these types of questions but this one has got me really stuck. I tried converting to polar coordinates but it gave a horrible integration to carry out for me so I think I've done it wrong. So yeah, any help appreciated (Happy).

I couldn't write out the integral on here so I've attached the paper that the question's from (it wouldnt let me attach a JPEG or GIF of the question - why?!). Question 10 (Happy).
 

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TheEmptySet

MHF Hall of Honor
Feb 2008
3,764
2,029
Yuma, AZ, USA
I'm normally ok with these types of questions but this one has got me really stuck. I tried converting to polar coordinates but it gave a horrible integration to carry out for me so I think I've done it wrong. So yeah, any help appreciated (Happy).

I couldn't write out the integral on here so I've attached the paper that the question's from (it wouldnt let me attach a JPEG or GIF of the question - why?!). Question 10 (Happy).
You need to check your algebra!

You have the integral

\(\displaystyle \oint P(x,y)dx+Q(x,y)dy\)

If you take

\(\displaystyle \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2x^4y+2y^5+4x^2y^3\)

Then you just need to integrate over the quarter circle in the first quadrant with radius a.
 
Jan 2011
14
0
Newcastle
Ahh whoops! Ok, so I checked and I had made a stupid mistake but I carried on with \(\displaystyle 4x^2y^3+2x^4y+2y^5\) in the integrand and converted to polars
\(\displaystyle x=acos\theta\)
\(\displaystyle y=asin\theta\)
\(\displaystyle dxdy=adad\theta\)
but I ended up with

\(\displaystyle 1/7 \pi a^7(2cos^2asin^3a+cos^4asina+sina)\)

which I'm pretty sure must be wrong. I'm not sure if I'm just being stupid or if I've been awake too long but if you can see where I've gone wrong, could you point it out for me? Thanks!
 

TheEmptySet

MHF Hall of Honor
Feb 2008
3,764
2,029
Yuma, AZ, USA
Ahh whoops! Ok, so I checked and I had made a stupid mistake but I carried on with \(\displaystyle 4x^2y^3+2x^4y+2y^5\) in the integrand and converted to polars
\(\displaystyle x=acos\theta\)
\(\displaystyle y=asin\theta\)
\(\displaystyle dxdy=adad\theta\)
but I ended up with

\(\displaystyle 1/7 \pi a^7(2cos^2asin^3a+cos^4asina+sina)\)

which I'm pretty sure must be wrong. I'm not sure if I'm just being stupid or if I've been awake too long but if you can see where I've gone wrong, could you point it out for me? Thanks!
I can confirm that your answer is incorret but you didn't show how you got it so I have no idea what you did wrong.

What I can tell you is that after you convert to polar coordinates you will have a double integral of the from

\(\displaystyle \int_{0}^{a}\int_{0}^{\frac{\pi}{2}}f(r, \theta)r d \theta dr\)

In you integrate with respect to theta first you should end up with


\(\displaystyle \int_{0}^{a} 2r^6dr\)
 
Jan 2011
14
0
Newcastle
I just can't get that (Crying) I don't know where I'm going wrong either!

Ok so changing to polar coordinates I get

\(\displaystyle 2r^6sin\theta(2cos^2\theta sin^2\theta+cos^4\theta+sin^4\theta)\)

in the integrand but I just can't integrate all the sine and cosines when integrating with respect to \(\displaystyle \theta\). Am I doing the change to polar coordinates wrong or something?
 

TheEmptySet

MHF Hall of Honor
Feb 2008
3,764
2,029
Yuma, AZ, USA
I just can't get that (Crying) I don't know where I'm going wrong either!

Ok so changing to polar coordinates I get

\(\displaystyle 2r^6sin\theta(2cos^2\theta sin^2\theta+cos^4\theta+sin^4\theta)\)

in the integrand but I just can't integrate all the sine and cosines when integrating with respect to \(\displaystyle \theta\). Am I doing the change to polar coordinates wrong or something?
You need to use the Pythagorean identity from trig.

\(\displaystyle 2\cos(\theta)\sin(\theta)+\cos^4(\theta)+\sin^4( \theta )=\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)=\)

\(\displaystyle \cos^2(\theta)[\cos^2(\theta+\sin^2(\theta))] + \sin^2(\theta)[\cos^2(\theta+\sin^2(\theta))]=\cos^2(\theta)+\sin^2(\theta)=1 \)

So you end up with

\(\displaystyle 2r^6\sin(\theta)\)
 
Jan 2011
14
0
Newcastle
Genius, thank you!! I think I just got scared by all the sines and cosines, thanks for all your help =]