How to determine bounds of polar graph without graphing?

Question

Question: Find the area in the second quadrant bounded by $r = \cos θ + \sin θ.$

answer with picture

Answer: The radial lines that bound the portion in the second quadrant are $θ = π/2$ (which makes sense) and $θ = 3π/4$. The second value makes $r = θ$ (since the sine and cosine of $3π/4$ are opposites), which is the intersection at the pole marking the end of the region’s presence in the second quadrant.

I was able to determine that $r=0$ in the second quadrant at $θ=3π/4$, but how do I know if the other bound is $π$ or $π/2$?

Answer 1

Curve is $r = \cos θ + \sin θ$. In polar coordinates, $x = r \cos \theta, y = r \sin \theta$.

Note that in second quadrant, $x \leq 0, y \geq 0$.

For $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{4}, |\sin\theta| \geq |\cos\theta|$, $\cos\theta$ is negative and $\sin\theta$ is positive.

So, $r \geq 0, x ( = r \cos\theta) \leq 0, y ( = r \sin\theta) \geq 0 $ and it is clear that the curve is in second quadrant.

For $\frac{3\pi}{4} \leq \theta \leq \pi, |\sin\theta| \leq |\cos\theta|$, again $\cos\theta$ is negative and $\sin\theta$ is positive.

So, $r \leq 0, x ( = r \cos\theta) \geq 0, y ( = r \sin\theta) \leq 0 $ and it is clear that the curve is in fourth quadrant.

Does that explain why $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{4}$ in second quadrant?

Answer 2

By tabulating without graphing we have corresponding polar coordinates

$$\theta (0, \pi/4,\pi/2, 3 \pi/4, \pi, 5 \pi/4, 3 \pi/2, 7 \pi/4 ) \;\&\; r (1, \sqrt 2, 1, 0,-1, -\sqrt 2, -1,0 ) $$

which gives the full calculated picture It is noted that in the second quadrant radius vector can be positive, negative or zero.

In this quadrant the total area is in fact zero.

The graphing ( rough sketch shown ) should be absent fully or be present in full, but not one half of it included up to the middle of domain.