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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!

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this seems to be missing the information about y stopping at 1... that seems like an issue in the setup... – Timtam Aug 3 '12 at 6:40
can anyone confirm if the rest of my technique was correct despite the initial mistake? – Timtam Aug 3 '12 at 6:46
There should be a $4\pi$ instead of the $2\pi$ in the numerator. You seem to have dropped a factor of $2$ somewhere. Edit: Observe that one of the terms in your final integral is $4\int_0^{\pi/3} \frac12\,\mathrm d\theta$, which is clearly $4\pi/6$. – Rahul Aug 3 '12 at 6:49
i think i see the problem its in the last step but otherwise i think it's right – Timtam Aug 3 '12 at 6:52
i'm finding it a bit tricky setting up the correct start and endpoints on some of these multiple integrals, but I think i understand this problem now, at least. – Timtam Aug 3 '12 at 7:05
up vote 3 down vote accepted

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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interesting i wasn't thinking about it this way i was thinking geometrically about the line y = 1 rather. but a line is a function! – Timtam Aug 3 '12 at 6:44
Maybe it would also be useful to compare the result with the formula of the area of circular segment. – Martin Sleziak Aug 3 '12 at 6:47
I am the habit to use parentheses when the function to integrate is a sum. Is considered a correct formalism also without them? – enzotib Aug 3 '12 at 10:10
I've edited both integrals containing difference the way you suggested @enzotib. I am not sure whether both forms are considered standard; this way is certainly clearer. – Martin Sleziak Aug 3 '12 at 10:16

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