Find the area bounded by the following curves: x²+y²=2x
Find the area bounded by the following curves: $$ x^2+y^2=2x, \qquad y=0, \qquad y=x. $$
Explanation of Method
Completing the square shows the first curve is a circle through the origin, which converts very cleanly to polar coordinates ($r=2\cos\theta$). The region cut out between the $x$-axis and the ray $y=x$ is then simply the circular sector $0\leqslant\theta\leqslant\frac\pi4$, $0\leqslant r\leqslant2\cos\theta$.
- Identify the circle by completing the square, and convert it to polar form.
- Identify the angular range cut out by $y=0$ and $y=x$.
- Integrate the resulting circular sector in polar coordinates.
Step 1 · Identify the circle
Completing the square in $x^2+y^2=2x$:
$$ (x-1)^2+y^2 = 1, $$
a circle of radius $1$ centered at $(1,0)$, passing through the origin. In polar coordinates ($x=r\cos\theta$), $x^2+y^2=2x$ becomes $r^2=2r\cos\theta$, i.e. $r=2\cos\theta$.
Step 2 · Identify the angular range
The ray $y=0$ (with $x\geqslant0$) corresponds to $\theta=0$; the ray $y=x$ (in the first quadrant) corresponds to $\theta=\dfrac\pi4$. The bounded region is the circular sector swept between these two rays, with $r$ running from $0$ out to the circle $r=2\cos\theta$:
$$ 0\leqslant\theta\leqslant\frac\pi4, \qquad 0\leqslant r\leqslant2\cos\theta. $$
Step 3 · Set up and evaluate the area integral
$$ A = \int_0^{\pi/4}\int_0^{2\cos\theta} r\,dr\,d\theta = \int_0^{\pi/4}\left[\frac{r^2}2\right]_0^{2\cos\theta}d\theta = \int_0^{\pi/4} 2\cos^2\theta\,d\theta. $$
Using $\cos^2\theta=\dfrac{1+\cos2\theta}2$:
$$ A = \int_0^{\pi/4}(1+\cos2\theta)\,d\theta = \left[\theta+\frac{\sin2\theta}2\right]_0^{\pi/4} = \frac\pi4+\frac12. $$
Additional Notes
- Verified numerically: direct quadrature of $\int_0^{\pi/4}2\cos^2\theta\,d\theta$ gives $\approx1.285398$, matching $\frac\pi4+\frac12\approx1.285398$.
Result
The area is: