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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!Quote:
Originally Posted by Natalie11391
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.
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.Quote:
Originally Posted by Natalie11391
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
$\displaystyle \int_{0}^{a} 2r^6dr$
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.Quote:
Originally Posted by Natalie11391
$\displaystyle 2\cos(\theta)\sin(\theta)+\cos^4(\theta)+\sin^4( \theta )=\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\cos ^2(\theta)\sin^2(\theta)+\sin^4(\theta)=$
$\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
$\displaystyle 2r^6\sin(\theta)$
Genius, thank you!! I think I just got scared by all the sines and cosines, thanks for all your help =]
