calculusmultivariable-calculuspolar coordinates
Find the area enclosed by $r = 1 + \sin\theta$ and $r = 1 – \sin\theta$
So, the curves are given by the following parametrizations:
$$ f_1(\theta) = ((1 + \sin \theta) \cos \theta,(1 + \sin \theta) \sin \theta)$$
$$ f_2(\theta) = ((1 – \sin \theta) \cos \theta,(1 – \sin \theta) \sin \theta)$$
It looks logical that I have to find the intersections.
How can I find the integral enclosed by the curves?
Fleshing out Aneesh's comment: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{2}r^2d\theta=-2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1-\sin\theta\right)^2\,\overbrace{d(1-\sin\theta)}^{-\cos\theta}=\left.-\frac{2}{3}\left(1-\sin\theta\right)^3\right|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=$$ $$=-\frac{2}{3}\left[(1-1)^3-(1+1)^3\right]=\frac{16}{3}$$
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
Best Answer
You can use symmetry and multiply $0.5\int_0^{\pi/2} (1-\sin\theta)^2 d\theta$ by $4$. When you expand, you get three integrals that are all standard.That I want you to try. You should find $\frac{3\pi-8}{2}$ (And not zero!)