FP3 Polar Co-ordinates Integral
Maths and statistics discussion, revision, exam and homework help.
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FP3 Polar Co-ordinates IntegralSomebody Please Rep Nudoi For Me !!
The curve
can be found on Q4B, page 139 here: http://store.aqa.org.uk/qual/pdf/AQA-MFP3-TEXTBOOK.PDF
I now need to find the area of it between
. Since it has two loops of equal area, I found the area of one loop and multiplied by two. The region for which it's defined for the loop in the first quadrant is
hence my integral is:
![\displaystyle 2 \times \dfrac{1}{2} \int^{\frac{\pi}{2}}_0 a^{2}\sin^2 2\theta \ d\theta
\
\
=\displaystyle \int^{\frac{\pi}{2}}_0 a^{2} \times \dfrac{1}{2} (1-\cos 4\theta) \ d\theta
\
\
= \left[\dfrac{\theta a^2}{2} + \dfrac{\sin 4\theta a^2}{8} \right]_0^{\frac{\pi}{2}} = \dfrac{\pi a^2}{4}](http://www.thestudentroom.co.uk/latexrender/pictures/2b/2bcc09be1c689b575dd5b22f5816db91.png "\displaystyle 2 \times \dfrac{1}{2} \int^{\frac{\pi}{2}}_0 a^{2}\sin^2 2\theta \ d\theta
\
\
=\displaystyle \int^{\frac{\pi}{2}}_0 a^{2} \times \dfrac{1}{2} (1-\cos 4\theta) \ d\theta
\
\
= \left[\dfrac{\theta a^2}{2} + \dfrac{\sin 4\theta a^2}{8} \right]_0^{\frac{\pi}{2}} = \dfrac{\pi a^2}{4}")
I have two questions and would appreciate it if someone could answer both:
1.
Can anyone think of a way of integrating this other than writing
, is there any sub that you can think of or re-arrangement ?
2. The book says
, who is correct ? I can't see any errors I have made and perhaps they have only found the area for one of the loops.
Thnx
Edit, If you want to look at the original question then it is Q4 on page 49 in the pdf above but the diagram for that particular curve can be found as the answer to another question which can be found on Q4B, P139.Last edited by member910132; 14-05-2012 at 22:20. -
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Re: FP3 Polar Co-ordinates Integral1. Not really, no. Think about it: if you are to have something which differentiates to give
then it must be of the form 
for some constant 
. Since this is a polynomial plus a trig function, you're unlikely to get any kind of 'nice' substitution without using a trig identity. Exploiting the double-angle formula really is the easiest way.
2. The question asks you to find the area enclosed in each of the loops, not in both loops together. So you're right in saying that the area enclosed by both loops is
, but that's not what you're asked for.
