Area Inside a Polar Curve

The sector is $\frac{1}{4}$ of the disk: $\frac{1}{4} \cdot 9\pi = \frac{9\pi}{4}$, which is $\frac{1}{2}r^2\theta = \frac{1}{2} \cdot 9 \cdot \frac{\pi}{2}$.

• Correct: $\frac{9\pi}{4}$:
• $\frac{3\pi}{2}$:
• $\frac{9\pi}{2}$:
• $9\pi$:

Thin pieces of size $f(x_i)\,\Delta x$, added up, as the pieces get thinner.

• Correct: Riemann sums $\sum f(x_i)\,\Delta x$:
• slopes of secant lines:
• the values $f(a)$ and $f(b)$:
• averages of $f(a)$ and $f(b)$:

$\sin(2\theta) = 0$ when $2\theta$ is a multiple of $\pi$.

• Correct: $0$, $\frac{\pi}{2}$ and $\pi$:
• $\frac{\pi}{4}$ and $\frac{3\pi}{4}$:
• $0$ and $\pi$:
• $\frac{\pi}{2}$:

The circle is traced once for $0 \le \theta \le \pi$: $\frac{1}{2}\int_0^{\pi} 16\sin^2\theta\,d\theta$$= 8 \cdot \frac{\pi}{2}$$= 4\pi$, the area of a circle of radius 2.

• Answer: $4\,\pi$ ≈ 12.566

$1 + 2\cos\theta = 0$ at $\theta = \frac{2\pi}{3}$ and $\frac{4\pi}{3}$. Between them $r < 0$, and the curve traces the inner loop once; squaring $r$ makes the integrand positive. The area is $\pi - \frac{3\sqrt{3}}{2}$.

• $\frac{1}{2}\int_{0}^{2\pi/3} (1 + 2\cos\theta)^2\,d\theta$: From 0 to $\frac{2\pi}{3}$ the curve traces the upper half of the big outer loop.
• $\frac{1}{2}\int_{-2\pi/3}^{2\pi/3} (1 + 2\cos\theta)^2\,d\theta$: That is the whole outer loop, where $r > 0$.
• Correct: $\frac{1}{2}\int_{2\pi/3}^{4\pi/3} (1 + 2\cos\theta)^2\,d\theta$: Right: $r < 0$ between the two zeros $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$, and that stretch traces the inner loop.
• $\frac{1}{2}\int_{0}^{2\pi} (1 + 2\cos\theta)^2\,d\theta$: A full turn counts the inner loop inside the outer one a second time.

A slice of a circle of radius $r$ with angle $\Delta\theta$ is the fraction $\frac{\Delta\theta}{2\pi}$ of the disk, so its area is $\frac{\Delta\theta}{2\pi}\pi r^2 = \frac{1}{2}r^2\,\Delta\theta$. Adding slices and taking the limit gives $\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$.

• Correct: A thin wedge is nearly a circle slice, of area $\frac{1}{2}r^2\,\Delta\theta$: Right: the polar area integral is the limit of these sectors' areas added up.
• It is the area of a thin rectangle of height $r$ and width $\Delta\theta$: Polar regions are cut into wedges, not rectangles; a wedge's area grows with $r^2$.
• It is the average of $r$ and $r^2$ over the interval: $\frac{1}{2}r^2$ is not an average; it is the slice area $\frac{\Delta\theta}{2\pi} \cdot \pi r^2$ per unit of $\Delta\theta$.
• The curve is traced twice, so the area is halved: Some curves are traced twice over $[0, 2\pi]$, which is a reason to change the limits, not the source of the $\frac{1}{2}$.