Find the area S bounded by the curves y^2 - 2y + x^2 = 0, y^2 - 4y + x^2 = 0, x = 0 and y = x*sqrt(3).

Step-by-step solution

1. 1

   Recognise the two curves as circles. Completing the square, $x^2+y^2-2y=0$ becomes $x^2+(y-1)^2=1$: a circle of radius $1$ centred at $(0,1)$. Likewise $x^2+y^2-4y=0$ becomes $x^2+(y-2)^2=4$: radius $2$, centre $(0,2)$. Both pass through the origin and are tangent to the $x$-axis there.

2. 2

   Convert to polar coordinates. With $x=r\cos\theta$, $y=r\sin\theta$ we get $r^2-2r\sin\theta=0\Rightarrow r=2\sin\theta$, and $r^2-4r\sin\theta=0\Rightarrow r=4\sin\theta$. Circles through the origin have exactly this clean polar form, which is why polar is the right coordinate system here.

3. 3

   Convert the two lines into angles. $x=0$ is the ray $\theta=\tfrac{\pi}{2}$. The line $y=x\sqrt3$ has $\tan\theta=\sqrt3$, so $\theta=\tfrac{\pi}{3}$. The region is therefore the annular sector $\tfrac{\pi}{3}\le\theta\le\tfrac{\pi}{2}$, $2\sin\theta\le r\le 4\sin\theta$.

4. 4

   Set up the polar area integral. $S=\displaystyle\int_{\pi/3}^{\pi/2}\!\!\int_{2\sin\theta}^{4\sin\theta} r\,dr\,d\theta=\int_{\pi/3}^{\pi/2}\frac{(4\sin\theta)^2-(2\sin\theta)^2}{2}\,d\theta=6\int_{\pi/3}^{\pi/2}\sin^2\theta\,d\theta$.

5. 5

   Use the power-reduction identity. $\sin^2\theta=\tfrac{1-\cos 2\theta}{2}$, so $\displaystyle\int\sin^2\theta\,d\theta=\frac{\theta}{2}-\frac{\sin 2\theta}{4}$.

6. 6

   Evaluate at the limits. At $\theta=\tfrac{\pi}{2}$: $\tfrac{\pi}{4}-0=\tfrac{\pi}{4}$. At $\theta=\tfrac{\pi}{3}$: $\tfrac{\pi}{6}-\tfrac{\sin(2\pi/3)}{4}=\tfrac{\pi}{6}-\tfrac{\sqrt3}{8}$. The difference is $\tfrac{\pi}{12}+\tfrac{\sqrt3}{8}$, and multiplying by $6$ gives $S=\tfrac{\pi}{2}+\tfrac{3\sqrt3}{4}$.

7. 7

   Check numerically. $\tfrac{\pi}{2}+\tfrac{3\sqrt3}{4}=1.570796+1.299038=2.869834$, and a Simpson evaluation of $6\int_{\pi/3}^{\pi/2}\sin^2\theta\,d\theta$ returns $2.869834$, matching to six decimals.