## anonymous one year ago hello i wonder how you can decide the angle when changing to polar coordinates? i know this: x=rcos(delta) y=rsin(delta) delta= arctan (x/y) but i find it very difficult to know the value of delta

1. freckles

can you give an example of where you have trouble at

2. freckles

|dw:1437432457639:dw| arctan will output values for these quadrants where theta is between -pi/2 and pi/2

3. freckles

so to get the other quadrants we have to be a bit more clever but not too much more

4. freckles

|dw:1437432533105:dw| that was a straight line so if you really meant to have the angle there you can adjust your theta by adding pi to it

5. freckles

|dw:1437432587064:dw| same thing for that 3rd quadrant you will find the angle in the 1st and you add pi to get the one in the 3rd if that is the one you are meant to have

6. anonymous

7. anonymous

i understand everything except the pi

8. freckles

Examples: EX 1: $(x,y)=(1,1) \\ r=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2} \\ \theta=\arctan(\frac{y}{x})=\arctan(\frac{1}{1})=\arctan(1)=\frac{\pi}{4} \\ (r,\theta)=(\sqrt{2},\frac{\pi}{4}) \\$ EX 2: $(x,y)=(-1,1) \\ r=\sqrt{(-1)^2+1^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2} \\ \theta=\arctan(\frac{y}{x})=\arctan(\frac{-1}{1})=\arctan(-1) \\ \text{ this will give the angle in the 4th quadrant } \\ \text{ you want the one in the 2nd } \\ \text{ do } + \pi \\ \theta=\arctan(\frac{y}{x})=\arctan(-1)+\pi=-\frac{\pi}{4}+\pi=\frac{3\pi}{4} \\ (r, \theta)=(\sqrt{2}, \frac{3\pi}{4})$ EX 3: $(x,y)=(-1,-1) \\ r=\sqrt{2} \\ \theta=\arctan(1) \text{ this will give you the one in the first quadrant } \\ \text{ we want the one in the 3rd } \\ \text{ do } +\pi \\ \theta=\arctan(1)+\pi =\frac{\pi}{4}+\pi=\frac{5\pi}{4} \\ (r,\theta)=(\sqrt{2},\frac{5\pi}{4})$ EX 4: $(x,y)=(1,-1) \\ r=\sqrt{2} \\ \theta=\arctan(-1)=\frac{-\pi}{4} \\ (r, \theta)=(\sqrt{2},\frac{-\pi}{4})$

9. freckles

10. freckles

|dw:1437433116823:dw|

11. freckles

where R is |dw:1437433154036:dw|

12. freckles

so are you asking how they got the limits ?

13. anonymous

the limits of the angle yes

14. freckles

do you know that looking at a quarter of a circle we would be looking from theta=0 to theta=pi/2 assuming we want the part in the 1st quadrant ?

15. freckles

|dw:1437433287671:dw|

16. freckles

|dw:1437433324784:dw|

17. freckles

|dw:1437433373914:dw|

18. freckles

that is what they have in your picture the top part of the circle

19. freckles

that is why they have theta is between 0 and pi

20. freckles

|dw:1437433484377:dw|

21. freckles

22. anonymous

but how do you see that it is in the second quadrant? and equals to pi?

23. freckles

well I drew it above

24. freckles

25. freckles

they gave us that we are suppose to look at these half top parts of x^2+y^2=1 and x^2+y^2=4

26. anonymous

okay i see, it's quite confusing in my opinion . I'm going to try some more examples. you are very polite for putting your time in my question, i really appreciate that!

27. freckles

I could reword the question Say it say to evaluate that same thing except R is the region in the first quadrant bounded by the circles x^2+y^2=1 and x^2+y^2=4 well here theta would be between 0 and pi/2

28. freckles

Well not reword the question but give a slightly different question by rewording it

29. freckles

anyways I hope that did help good luck

30. anonymous

yes it's getting clearer and clearer now. thanks again!