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For the polar curve $r=2\cos(3\theta)$, you can find the area of all three petals by getting the area of one-half petal using the bounds $\theta = 0$ to $\theta = \pi$/6 for $A = (1/2)\int r^2 d\theta$, and multiplying that by 6. But when you try to get the area of the curve using the bounds $\theta = 0$ to $\theta = 2\pi$, the area is incorrect. My reasoning for why the latter method should work is that $A = (1/2)\int r^2 d \theta$, which will be the area within the bounds set by the integral. So why would the integral with the bounds $0$ to $2\pi$ not yield the correct area? The graph in the picture is for $r=\cos(3\theta)$.

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While you haven't shown your computation, if you did it correctly you should have found that your integral from $0$ to $2\pi$ came out as twice the correct answer. The reason is that as $\theta$ increases from $0$ to $2\pi$, the rose curve is drawn twice. So you should only integrate from $0$ to $\pi$ to get the correct value. Note that $\pi$ is $6\cdot \pi/6$-this isn't coincidental. Each $\pi/6$ increase in $\theta$ draws half of a petal.

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The reason is because when $\cos 3\theta < 0$, then the radius becomes negative. On the interval $\theta \in [0,2\pi)$, this occurs when $$\theta \in (\pi/6, \pi/2) \cup (5\pi/6, 7\pi/6) \cup (3\pi/2, 11\pi/6).$$ So in fact, this curve is traversed twice as $\theta \in [0, 2\pi)$.

Refer to the following diagram. As $\theta$ traverses the interval, when $r > 0$, the radial line with angle $\theta$ is blue and the point it is drawing is also blue. When $r < 0$, the line and point becomes red. The black circle represents the angle of the line, so as you can see, the line only makes a single complete rotation, but the curve is drawn twice.

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