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Find the area $\frac{\pi}{4}\leq\theta \leq\frac{\pi}{2}\;\;\;\;,0\leq r\leq2$

using double integral

Attempt:

I tried in two differnt ways and got two diffrent answers, can't find my mistakes

first methode: poolar coordinates

$$\int_{\pi/4}^{\pi/2}\int _0^2rdrd\theta=...=\pi/2$$

which is obvios wrong

seconde methode

$$\int_{0}^{\sqrt 2}\int_{x/2}^{x}dydx+\int_{\sqrt 2}^{4/\sqrt 5}\int_{x/2}^{\sqrt{4-x^2}}dydx=...\approx 1/2+0.14$$

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    $\begingroup$ The area in polar coordinates is given by $$\frac{1}{2}\int_{\theta_0}^{\theta_1}r(\theta)^2\,d\theta.$$ $\endgroup$ – Jack D'Aurizio Aug 2 '16 at 18:53
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    $\begingroup$ @JackD'Aurizio when he performs the first integration, it'll become the same $\endgroup$ – GeorgSaliba Aug 2 '16 at 18:54
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    $\begingroup$ So $\frac{\pi}{2}$ is obviously right, since the area of the eighth part of a circle with radius $2$ is $\frac{4\pi}{8}=\frac{\pi}{2}$. $\endgroup$ – Jack D'Aurizio Aug 2 '16 at 18:55
  • $\begingroup$ @Error404 $\theta = \pi/2$ is at $90^\circ$. So where are your bounds coming from in your second method? $\endgroup$ – user137731 Aug 2 '16 at 18:55
  • $\begingroup$ I made a sketch @Bye_World $\endgroup$ – Error 404 Aug 2 '16 at 18:58
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Your bounds in the second method are wrong. They should be $$\int_{0}^{\sqrt{2}}\int_{x}^{\sqrt{4-x^2}}dydx$$

Here's what a plot of the region should look like (easy to plot from polar coordinates).

Hold up a minute and I'll graph it to make it easier to understand where these came from.

So my approach here was to first get an inequality of $y$ in terms of $x$. So coming up from the $x$-axis, we first enter the region at the line $y=x$ (the $\pi/4$ radian line). Then we leave one we get to the circle $r=2$, or equivalently $x^2+y^2=4$. Thus I get the bounds on $y$.

Then $x$ is easy. We just take the lowest and highest values of $x$ in the region. Clearly $x$ starts at $0$ and then moves along until the very point where its on the $r=2$ circle at $\pi/4$ rad. That's clearly at $2\cos(\pi/4) = \sqrt{2}$.

Thus the Cartesian bounds are obtained.

Note that this exercise should have been really easy to figure out geometrically. Your region is just one eighth of a circle and thus its area is $$A = \frac{1}{8}(\pi(2)^2) = \frac{\pi}{2}$$

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In the first quadrant, the area is delimited by the arc $y=\sqrt{4-x^2}$ from the top, the line $y=x$ from the bottom, $x=0$ to the left and $x=2\cos(\dfrac \pi 4)$ to the right.

Alternatively, you can integrate $y=\sqrt{4-x^2}$ between $0$ and $\sqrt 2$ then subtract the area of the triangle $A=\frac 12\sqrt 2 \sqrt 2=1$

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$$I=\int_{0}^{\sqrt 2}\int_{x}^{\sqrt{2}}dydx+\int_{0}^{\sqrt 2}\int_{\sqrt 2}^{\sqrt{4-x^2}}dydx,$$ $$I=\int_{0}^{\sqrt 2}(\sqrt{2}-x)dx+\int_{0}^{\sqrt 2}(\sqrt{4-x^2}-\sqrt 2)dx.$$ $$I=-1+\int_{0}^{\sqrt 2}\sqrt{4-x^2}dx=-1+J.$$ Now use substitution $x=2\sin{t}$ to solve integral $J$.

It is easy to calculate $$J=1+\frac{\pi}{2}.$$ Therefore, $$I=\frac{\pi}{2}.$$ The same result you will obtain if you use polar coordinates.

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