Polar coordinates — practice questions

This is a cardioid. The presence of $\cos\theta$ implies symmetry about the initial line ($"x"$-axis). We can sketch for $0 \le \theta \le \pi$ and then reflect.

1. Table of Values: | $\theta$ | $0$ | $\pi/3$ | $\pi/2$ | $2\pi/3$ | $\pi$ | |---|---|---|---|---|---| | $\cos\theta$ | $1$ | $1/2$ | $0$ | $-1/2$ | $-1$ | | $r$ | $6$ | $4.5$ | $3$ | $1.5$ | $0$ |
2. Key Features:
   • Maximum r: Occurs when $\cos\theta = 1$, so $\theta=0$. Max $r = 3(1+1) = 6$. The point is $(6, 0)$ in Cartesian coordinates.
   • Minimum r: Occurs when $\cos\theta = -1$, so $\theta=\pi$. Min $r = 3(1-1) = 0$. The curve passes through the pole at $\theta=\pi$.
   • Tangent at the pole: The tangent at the pole is the line $\theta = \pi$ (the negative x-axis).
   • When $\theta = \pi/2$, $r=3$. The point is $(0, 3)$ in Cartesian coordinates.
3. Sketching:
   • Start at $\theta=0$, where the curve is at its furthest point from the pole, $(6,0)$.
   • As $\theta$ increases to $\pi/2$, $r$ decreases from $6$ to $3$. The curve moves inwards, passing through $(0,3)$.
   • As $\theta$ increases from $\pi/2$ to $\pi$, $r$ decreases from $3$ to $0$, arriving at the pole with a tangent along the line $\theta=\pi$.
   • Use symmetry to draw the lower half of the curve for $\pi < \theta < 2\pi$. The curve leaves the pole and returns to the starting point $(6,0)$. The resulting shape is a cardioid (heart-shape) on its side.

The curve $r = 4\sin(2\theta)$ is a four-petalled rose. The given range $0 \le \theta \le \pi/2$ corresponds to the first petal in the first quadrant.

1. Set up the integral: The area $A$ is given by the formula $A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta$. Here, $r = 4\sin(2\theta)$, $\alpha=0$, and $\beta=\pi/2$. $A = \frac{1}{2} \int_{0}^{\pi/2} (4\sin(2\theta))^2 \, d\theta$ $A = \frac{1}{2} \int_{0}^{\pi/2} 16\sin^2(2\theta) \, d\theta$ $A = 8 \int_{0}^{\pi/2} \sin^2(2\theta) \, d\theta$
2. Use a trigonometric identity: We use the double angle identity $\cos(2A) = 1 - 2\sin^2(A)$, which rearranges to $\sin^2(A) = \frac{1}{2}(1 - \cos(2A))$. Let $A = 2\theta$, so $\sin^2(2\theta) = \frac{1}{2}(1 - \cos(4\theta))$. $A = 8 \int_{0}^{\pi/2} \frac{1}{2}(1 - \cos(4\theta)) \, d\theta$ $A = 4 \int_{0}^{\pi/2} (1 - \cos(4\theta)) \, d\theta$
3. Integrate and evaluate: $A = 4 \left[ \theta - \frac{1}{4}\sin(4\theta) \right]_{0}^{\pi/2}$ $A = 4 \left( (\frac{\pi}{2} - \frac{1}{4}\sin(2\pi)) - (0 - \frac{1}{4}\sin(0)) \right)$ Since $\sin(2\pi) = 0$ and $\sin(0) = 0$, this simplifies to: $A = 4 \left( \frac{\pi}{2} - 0 - 0 \right)$ $A = 2\pi$

The area of one petal is $2\pi$ square units.