Converting iterated integral to polar coordinates

[Seimuna]

erm...im facing problem in identifying the limit of double integral in polar coordinates...
normal ll jus substitute with \(\displaystyle \int_{0}^{2/\pi} and \int_{0(why.always.zero.???)}^{2(r=2.???)}\)

\(\displaystyle \int_{-2}^{2} \int_{0}^{\sqrt{[4-x^{2}]}} [Sin[x^2+y^2]] dy dx\)

= \(\displaystyle \int_{0}^{2/\pi} \int_{0}^{2} [r Sin r^2] dr d{\theta}\)
= \(\displaystyle \int_{0}^{2/\pi} [-(1/2)(Cos[2^2])]-[-(1/2)(Cos[(0)^2])] d{\theta}\)
= \(\displaystyle \int_{0}^{2/\pi} [(1/2)-(1/2)(Cos[4])] d{\theta}\)
= \(\displaystyle (2\pi)(1/2)[1-cos[4]]\)
= \(\displaystyle [1-cos[4]]\pi\)

\(\displaystyle \int_{0}^{2} \int_{\sqrt{[4-x^{2}]}}^{0} y dy dx\)

= \(\displaystyle \int_{-\pi/2}^{\pi/2} \int_{???}^{0} r sin {\theta} r dr d{\theta}\)

\(\displaystyle \int_{0}^{2} \int_{y}^{-\sqrt{[2y-y^{2}]}} x dx dy\)

=\(\displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{2 sin {\theta}} [r cos {\theta}] r dr d{\theta}\)

 

[tkhunny]

Polar coordinates are in terms of r and \(\displaystyle \theta\). Why are you puzzled about the limits being in terms of 'r'? If you are looking for an area that contains the Origin, it is likely that r = 0 is a good place to start.

Do you really mean \(\displaystyle 2/\pi\)? Maybe \(\displaystyle 2\pi\)?

It's not always clear that \(\displaystyle [0,2\pi]\) is the right plan. Are you SURE you traversed the curve only once?

It is ALWAYS wise to observe symmetries and exploit them. Your area is quite obviously the top half of a circle of radius 2 and your integrand has circular symmetries. Why not do \(\displaystyle [0,\frac{\pi}{2}]\) and multiply by 2?