<s>[INST] Calculating area between circle and line - where have I gone wrong here I think the initial set up is wrong below I should just integrate over the area as a multiple integral here. correct? the 1step below seems wrong.
Problem is asking for the area inside a region of a circle of radius 2 (centered on origin) and with x> 0 and y > 1. Here's my attempt at a solution
$A = \int_0^\sqrt3 \sqrt{4-x^2}dx $
since when x varies from $\sqrt3$ to 0 as y varies from 1 to the circles edge. So then I tried a substitution
$x= 2\sin\theta\qquad$
 $ dx = 2\cos\theta d\theta$
$A = \int_0^{\frac{\pi}{3}} \sqrt{4-(2sin\theta)^2}dx$
$A = 2\int_0^{\frac{\pi}{3}}  \cos\theta dx $
$A = 2\int_0^{\frac{\pi}{3}}  \cos\theta dx  = 4\int_0^{\frac{\pi}{3}}  \cos^2\theta d\theta = 4\int_0^{\frac{\pi}{3}} \frac{1}{2}\left( 1 +  \cos2\theta\right) d\theta$
Here I substituted for $2\theta = v$ so $dv = 2d\theta$; and the limits change by a factor of two on the second integral so it should be
$ A= \frac{\theta}{2}|^\frac{\pi}{3}_0 + \sin{v} |^\frac{2\pi}{3}_{0} = \frac{2\pi + 3\sqrt3}{6}  $
whereas the textbook gives the solution as = $\frac{4\pi -3\sqrt3}{6}$
I know the book answers is right since the total area of the quadrant is only pi but I'm not sure where exactly I'm screwing this up any tips appreciated!
 [/INST] If we use formula for the area of circular segment
$$2A = \frac{R^2}{2} \left(\theta - \sin\theta \right)$$ 
with $R=2$, $\theta = \frac{2\pi}3$ and $\sin\theta=\sin\frac{2\pi}3=\sin\frac\pi3=\frac{\sqrt3}2$, we get
$$2A=2\left(\frac{2\pi}3-\frac{\sqrt3}2\right)\\
A=\frac{2\pi}3-\frac{\sqrt3}2$$
which is the same result as in the book.

You need $A = \int_0^\sqrt3 (\sqrt{4-x^2} -1) \,dx = I-\sqrt 3$, where $I$ denotes the integral you have computed.
In general $\int_a^b (g(x)-f(x)) \,dx$ is area between two functions $f$ and $g$.
You also missed factor 2 in the last step in $2\int_0^{\pi/3} 1\, d\theta$, see Rahul Narain's comment.
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