<s>[INST] Using integration to compute the area of a region bounded by a curve For the area bounded by curves 
$$(x-1)^2 + y^2 =1 $$
and
$$x^2 + y^2 =1$$
I tried it by finding $y$ and integrating within the suitable limits through $y\,\mathrm{d}x$ method. But when I tried the $x\,\mathrm{d}y$ way the answer was different. For the $y\,\mathrm{d}x$ method the answer was $\frac{2\pi}{3} - \frac{\sqrt3}2$ . But the answer for $x\,\mathrm{d}y$ method was not same.
(y.dx means I found the function y in terms of x and integrated in the respective limits. x.dy means I found the function x in terms of y and integrated within respective limits.)
 [/INST] Through geometry, we can see that the area bound by the curves is made up of $4$ equal areas -- namely the area bound by the upper red semicircle and the $x$-axis on the interval $x\in[0.5,1]$.

If the shaded area above is $A$, then the total area bound by the curves is $4A$.
We can then write the red semicircle as a function $y$ in terms of $x$, and perform the integration.
$$x^2+y^2=1\Rightarrow y=\sqrt{1-x^2}$$
\begin{align}
A&=\int_{0.5}^1\sqrt{1-x^2}\ dx\\
4A&=4\int_{0.5}^1\sqrt{1-x^2}\ dx\\
&=4\int_{\frac{\pi}6}^{\frac{\pi}2}\sqrt{1-\sin^2\theta} \cos\theta\ d\theta&(\textrm{let }x=\sin\theta\Rightarrow dx=\cos\theta\ d\theta)\\
&=4\int_{\frac{\pi}6}^{\frac{\pi}2}\cos^2\theta\ d\theta\\
&=4\int_{\frac{\pi}6}^{\frac{\pi}2}\frac12(1+\cos 2\theta)\ d\theta\\
&=2\int_{\frac{\pi}6}^{\frac{\pi}2}1+\cos 2\theta\ d\theta\\
&=2(\theta+\frac12\sin 2\theta)\bigg]_{\frac{\pi}6}^{\frac{\pi}2}\\
&=2\theta+\sin 2\theta\bigg]_{\frac{\pi}6}^{\frac{\pi}2}\\
&=(\pi-\frac{\pi}3)+(\sin\pi-\sin\frac{\pi}3)\\
&=\frac{2\pi}3-\frac{\sqrt3}2
\end{align}
</s>