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!
Best Answer
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.