calculuspolar coordinates
Find the area of the region that is bounded by $r = \cos(\theta)$ from $0$ to $\pi/6$.
I'm pretty confident that I know what I am doing for this problem but I just want to make sure I have it down.
Evaluating the integral left me with $(1/4) [\theta + \sin(2\theta)/2]$ from $0$ to $\pi/6$
My answer is $\pi/24 + \sqrt{3}/16$
Could anyone confirm that I'm doing this right? I don't have any other way of checking really…
There are two distinct regions where your curves overlap. They do overlap for the intervals of $\theta$ that you give, but that gives only the large overlap at the upper right of the origin. You left out the smaller overlap at the lower left of the origin.
That overlap is covered by $\frac{5\pi}4\le \theta\le \frac{3\pi}2$ for the $r=1+\cos\theta$ curve and by $\pi\le \theta\le \frac{5\pi}4$ for the $r=1+\sin\theta$ curve. To get the correct answer you will need to include that little piece.
You want the region satisfying $1<r\le1+\sin\theta$, which implies $\sin\theta>0$ and $0<\theta<\pi$. So your calculation should be of$$\frac12\int_0^{\pi}[(1+\sin\theta)^2-1]d\theta=\int_0^\pi[\sin\theta+\tfrac12\sin^2\theta]d\theta=2+\frac{\pi}{4}.$$
Best Answer
the area in polar coordinate is: $$A=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}r^{2}(\theta)d\theta$$ then $$A=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}\cos^{2}(\theta)d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}\frac{1+\cos2\theta}{2}d\theta=\frac{1}{4}\left(\frac{\pi}{6}+\frac{\sin(\frac{\pi}{3})}{2}-\frac{\sin0}{2}\right)=$$ $$A=\frac{1}{8}\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)=\frac{\pi}{24}+\frac{\sqrt{3}}{16}$$ in conclution your answer is correct.
God bless you.