# Bounded area (again)

• May 29th 2009, 02:01 PM
Stonehambey
Bounded area (again)
Hi guys, sorry I'm not having much luck with polar coords today, there's something about them that makes me make a lot of mistakes then be unable to find them :P

Anyway, we're looking at the curves

$\displaystyle r=4\sin^2\theta$, $\displaystyle 0\leq\theta<2\pi$

$\displaystyle r=(2\sqrt{3})\sin2\theta$, $\displaystyle 0\leq\theta<\frac{\pi}{2}$

I have shown they intersect at (0,0) and (3, pi/3), and now I need to find the area of the region inside both the curves. Well after sketching them I came up with the area as being

$\displaystyle A = \frac{1}{2}\int^\frac{\pi}{2}_\frac{\pi}{3}(2\sqrt 3\sin2\theta)^2 \, d\theta + \frac{1}{2}\int^\frac{\pi}{3}_0(4\sin^2\theta)^2 \, d\theta$

After a page of working I arrive at

$\displaystyle 3\left[\theta-\frac{1}{4}\sin4\theta\right]^{\frac{\pi}{2}}_{\frac{\pi}{3}}+2\left[\theta-\sin2\theta+\frac{1}{8}\sin4\theta+\frac{\theta}{2 }\right]^{\frac{\pi}{3}}_0$

Which, when I punch into the calculator, gives me 2.11 (3 s.f.), when the answer in the back of the book gives 1.33

I'm really tenacious and it really bugs me when after so much working I get the wrong answer, especially when I go through the problem multiple times after and still can't see my error (Headbang)

Thanks :)
• May 30th 2009, 04:59 AM
Media_Man
A job well done...
Your integral is set up correctly. Your evaluation of that integral is correct. Your numerical value of the area is correct, $\displaystyle A\approx2.11431275$. Good job. Either you have miscopied the original problem or your text is simply wrong.