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I have a function:

$(a)$ $r = 4\cos(2\theta)$

$(b)$ $r = 4\sin(2\theta)$.

I need at least a set up for the integral that will yield the area inside the rose (a) but outside the rose $(b).$

I cant seem to figure out which strip to use because the boundaries are confusing me. Any help?

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First of all it's good to graph the functions $(a)$ (red) and $(b)$ (blue):

graph of a and b

The second step is to fund points of intersection, so we find all solutions of

$$\begin{cases} r=|4\cos(2\theta)|\\ r=|4\sin(2\theta)|\\ \end{cases}$$

We are taking absolute values because $r$ may be negative and this might complicate things. The solutions are:

$$r=2\sqrt2,\quad\theta\in\left\{\pm\tfrac18\pi,\pm\tfrac38\pi,\pm\tfrac58\pi,\pm\tfrac78\pi\right\}$$

It's enough to find area between $\theta=0$ and $\theta=\tfrac{\pi}{8}$ and multiply it by $8$:

$$S=8\cdot\left( \frac12\int_{0}^{\pi/8}r_a^2\,d\theta- \frac12\int_{0}^{\pi/8}r_b^2\,d\theta \right)$$

where $r_a=4\cos(2\theta),r_b=4\sin(2\theta)$. Simplifying:

$$S=8\cdot16\cdot\left( \frac12\int_{0}^{\pi/8}(\cos(2\theta))^2\,d\theta- \frac12\int_{0}^{\pi/8}(\sin(2\theta))^2\,d\theta \right)$$ $$=64\int_{0}^{\pi/8}(\cos(2\theta)^2-\sin(2\theta)^2)\,d\theta$$ $$=64\int_{0}^{\pi/8}\cos(4\theta)\,d\theta$$ $$=64/4\big[\sin(4\theta)\big]_0^{\pi/8}=16$$

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Note that in the first quadrant the two curves intersect at $\sin(2\theta)=\cos(2\theta) \Rightarrow \theta=\frac{\pi}{8}$.

In this quadrant: the curve (a) $r=4\cos(2\theta)$ starts from $(1,0)$ for $\theta=0$ and the curve (b) $r=4\sin (2\theta)$ starts at $(0,0)$. So the area inside (a) but not (b) is

$$ A_1=A_a-A_b=\frac{1}{2}\int_{0}^{\frac{\pi}{8}}\left(4\cos(2\theta)\right)^{\;2} d \theta-\frac{1}{2}\int_{0}^{\frac{\pi}{8}}\left(4\sin(2\theta)\right)^{\;2} d \theta=2 $$

enter image description here

where ( see the figure) $A_a$ is the area delimited by the blue arc $AP$ ( rose (a)), the segment $PO$,and the segment$OA$, and $A_b$ is the area delimited by the red arc $OP$ (rose (b)) and the segment $PO$.

Use the symmetry to find the entire area inside (a) but not (b) $A=8A_1=16$.

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