1. Polar Coordinates Definition, Conversions, and Integration

2. Where is it? In rectangular coordinates: We break up the plane into a grid of horizontal and vertical line lines. We locate a point by identifying it as the intersection of a vertical and a horizontal line. In polar coordinates: We break up the plane with circles centered at the origin and with rays emanating from the origin. We locate a point as the intersection of a circle and a ray. Coordinate systems are used to locate the position of a point. (3,1) (1,  /6 )

3. Locating points in Polar Coordinates Suppose we see the point and we know it is in polar coordinates. Where is it in the plane? (r,  )= (2,  /6 ) The first coordinate, r =2, indicates the distance of the point from the origin. (2,  /6 ) The second coordinate,  =  /6, indicates the distance counter- clockwise around from the positive x-axis. r =2  =  /6

4. Locating points in Polar Coordinates Note, however, that every point in the plane as infinitely many polar representations. (2,  /6 )  =  /6

5. Locating points in Polar Coordinates Note, however, that every point in the plane as infinitely many polar representations.

6. Locating points in Polar Coordinates Note, however, that every point in the plane as infinitely many polar representations. And we can go clockwise or counterclockwise around the circle as many times as we wish!

7. Converting Between Polar and Rectangular Coordinates It is fairly easy to see that if (x,y) and (r,  ) represent the same point in the plane: These relationships allow us to convert back and forth between rectangular and polar coordinates

8. Integration in Polar Coordinates Non-rectangular Integration Elements

9. Small Changes in r and  Suppose we consider a small change from r... to r + dr r r + dr

10. Small Changes in r and  Suppose we consider a small change from r... to r + dr This gives us a thin “ring” around the origin. r r + dr

11. Small Changes in r and  Suppose we consider a small change from ... to  + d    + d 

12. Small Changes in r and  Suppose we consider a small change from ... to  + d  This gives us a “pie-shaped wedge” that is subtended by the angle d    + d  dd

13. Small Changes in r and  Intersecting the “thin ring”... and the “pie-shaped wedge”,... we get...

14. Small Changes in r and  Intersecting the “thin ring”... and the “pie-shaped wedge”,... we get...

15. Small Changes in r and  dr dd In order to integrate a function given in polar coordinates (without first converting to rectangular coordinates!), we need to know the area of this little piece. Why?

16. Integration in Polar Coordinates In order to integrate a function given in polar coordinates, we will first “chop up” our region into a bunch of concentric circles and rays emanating from the origin. ( r *,  *) ( r *,  *, f(r*,  *)) Problem: the volume of the “tower” is the area of the base times the height. But the base is not a rectangle, so its area is not dr d  Now do this for each little “wedge” and add up the volumes of the “towers”.

17. Area of the “Small Bit”  A= area of sector of a circle

18. Area of a “Small Bit” dr dd In order to integrate a function given in polar coordinates (without first converting to rectangular coordinates!), we need to know the area of this little piece. r

19. Area of a “Small Bit” So...