How to find area of ​a circle? I have the equation in polar coordinates:
$$r=\cos(\phi)-\sin(\phi)$$
I just moved to 
Cartesian coordinates and drew a graph:
$$\left(x-\frac{1}{2}\right)^2+\left(y-\left(-\frac{1}{2}\right)\right)^2=\left(\sqrt{\frac{1}{2}}\right)^2$$

And now I should find the area of this circle using integral. What I need to do?Please help
 A: If you are going to do the algebra to change the coordinate system, you might as well finish the job algebraically.
$(x-h)^2 + (y-k)^2 = r^2$
and $A = \pi r^2 \implies A = \frac {\pi}{2}$
other options.
$r = \cos \phi - \sin \phi\\
r = \sqrt 2 (\cos (\phi + \frac {\pi}{4}))$
let $\theta = \phi + \frac {\pi}{4}$
$r = \sqrt 2 \cos \theta$
Which you might recognize as a circle of radius $\frac {\sqrt 2}{2}$
Caculus in polar coordinates
$A = \int_a^{b} \frac 12 r^2 \ d\phi$
The tricky part, $r\ge 0$ so you must find limits for $\phi$ such that this is true.
$\cos\phi \ge \sin\phi\\
\phi \in [-\frac {3\pi}{4},\frac {\pi}{4}]$
$\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \frac 12 (cos\phi - \sin\phi)^2\ d\phi\\
\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \frac 12 (cos^2\phi + \sin^2\phi - 2cos\phi\sin\phi)\ d\phi\\
\frac 12 (\phi + \frac 12 \sin^2 \phi)|_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}\\
\frac {\pi}{2}$
And finally integrate in Cartesian.
$(x-\frac {1}{2})^2 + (y+\frac {1}{2})^2 = \frac 12\\
y = \pm \sqrt {\frac 12 - (x-\frac {1}{2})^2} - \frac {\sqrt 2}{2}\\
\int_{\frac 12 - \frac {\sqrt 2}{2}}^{\frac 12 + \frac {\sqrt 2}{2}} (\sqrt {\frac 12 - (x-\frac {1}{2})^2} - \frac {\sqrt 2}{2}) - (-\sqrt {\frac 12 - (x-\frac {1}{2})^2} - \frac {\sqrt 2}{2})\ dx\\
\int_{\frac 12 - \frac {\sqrt 2}{2}}^{\frac 12 + \frac {\sqrt 2}{2}} 2\sqrt {\frac 12 - (x-\frac {1}{2})^2}\ dx\\
x - \frac 12 = \frac {\sqrt 2}{2} \sin t\\dx = \frac {\sqrt 2}{2}\cos t\ dt\\
\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \cos^2 t\ dt\\
\frac 12 (t + \sin t\cos t)|_{-\frac {\pi}{2}}^{\frac {\pi}{2}}\\
\frac {\pi}{2}$
A: lets take $$A=4\int_{0}^r \sqrt{r^2-x^2}dx$$
A: Another way, because of the given parametrization of the circle: using Green's Theorem with the parametrization
$$\begin{align*}&r(t)=\left(\frac12+\frac1{\sqrt2}\cos t\,,\,\,-\frac12+\frac1{\sqrt2}\sin t\right)\,,\\
&r'(t)=\left(-\frac1{\sqrt2}\sin t\,,\,\,\frac1{\sqrt2}\cos t\right)\end{align*}\;\;\;\;\;0\le t\le 2\pi$$
so:
$$A=\oint_C-y\,dx=\int_0^{2\pi}\left(\frac12-\frac1{\sqrt2}\sin t\right)\left(-\frac1{\sqrt2}\sin t\right)dt=\int_0^{2\pi}\left(-\frac1{2\sqrt2}\sin t+\frac12\sin^2t\right)dt$$
$$=\left.\frac1{2\sqrt2}\cos t\right|_0^{2\pi}+\left.\frac14\left(t-\cos t\sin t\right)\right|_0^{2\pi}=\frac\pi2$$
