A figure is bounded by the circles x² + y² − 7y = 0 and x² + y² − 9y = 0 and by the lines y = x/√3 and x = 0. Write the integral for its area in polar coordinates and find the area.

Step-by-step solution

1. 1

   Convert the two circles to polar form. Substituting $x=r\cos\varphi$, $y=r\sin\varphi$ turns $x^2+y^2$ into $r^2$, so $x^2+y^2-7y=0$ becomes $r^2-7r\sin\varphi=0$, i.e. $r(r-7\sin\varphi)=0$. Discarding the pole $r=0$ gives $r=7\sin\varphi$, and the same work on the second circle gives $r=9\sin\varphi$. Both are circles through the origin with centres on the $y$-axis, so in this region the radius runs from the smaller to the larger:

   $7\sin\varphi \le r \le 9\sin\varphi.$

2. 2

   Convert the two straight lines to angles. A line through the origin is a single ray $\varphi=\text{const}$ in polar coordinates. The line $x=0$ is the $y$-axis, so $\varphi=\tfrac{\pi}{2}$. For $y=\tfrac{x}{\sqrt3}$ substitute the polar forms: $r\sin\varphi=\tfrac{r\cos\varphi}{\sqrt3}$, so $\tan\varphi=\tfrac{1}{\sqrt3}$ and $\varphi=\tfrac{\pi}{6}$. The angle therefore sweeps

   $\frac{\pi}{6}\le\varphi\le\frac{\pi}{2}.$

3. 3

   Assemble the area integral. The area element in polar coordinates is $dS=r\,dr\,d\varphi$ — the extra factor $r$ is the Jacobian and is the single most common thing to forget here. With the limits found above,

   $S=\int_{\pi/6}^{\pi/2}\int_{7\sin\varphi}^{9\sin\varphi} r\,dr\,d\varphi.$

4. 4

   Do the inner integral in $r$. Because the integrand is just $r$, the inner integral is a difference of squares, which is why the $\sin^2\varphi$ factors out cleanly:

   $\int_{7\sin\varphi}^{9\sin\varphi} r\,dr=\frac{(9\sin\varphi)^2-(7\sin\varphi)^2}{2}=\frac{81-49}{2}\sin^2\varphi=16\sin^2\varphi.$

5. 5

   Integrate in $\varphi$ using the power-reduction identity. $\sin^2\varphi$ has no elementary antiderivative on sight, so rewrite it as $\sin^2\varphi=\tfrac{1-\cos 2\varphi}{2}$:

   $S=\int_{\pi/6}^{\pi/2}16\sin^2\varphi\,d\varphi=8\int_{\pi/6}^{\pi/2}(1-\cos 2\varphi)\,d\varphi=8\left[\varphi-\frac{\sin 2\varphi}{2}\right]_{\pi/6}^{\pi/2}.$

6. 6

   Evaluate at the limits. At $\varphi=\tfrac{\pi}{2}$: $\tfrac{\pi}{2}-\tfrac{\sin\pi}{2}=\tfrac{\pi}{2}$. At $\varphi=\tfrac{\pi}{6}$: $\tfrac{\pi}{6}-\tfrac{1}{2}\sin\tfrac{\pi}{3}=\tfrac{\pi}{6}-\tfrac{\sqrt3}{4}$. Subtracting and multiplying by $8$:

   $S=8\left(\frac{\pi}{2}-\frac{\pi}{6}+\frac{\sqrt3}{4}\right)=8\cdot\frac{\pi}{3}+2\sqrt3=\frac{8\pi}{3}+2\sqrt3\approx 11.84.$

   A numeric check of the same integral confirms $11.8417$, so the closed form is right.