Post History
#2: Post edited
by
r~~
·
2021-08-25T15:22:45Z (over 3 years ago)
- Regardless of the coordinate system, the principle of finding the limits of integration is the same: find the minimum and maximum value of the independent coordinate that enclose the region of interest.
- In polar coordinates, for these two particular curves, the region of interest is best described as two parts. As theta sweeps anticlockwise, first it covers the area between the origin and the blue curve, until reaching the (non-origin) point of intersection between the blue and red curves. Then it covers the area between the origin and the red curve, up until the red curve meets the origin again.
- (This description would need to be modified for different curves. In particular, if the line segment between the origin and a point on either curve ever intersected the curve at a third point, you would need to take that into account. You can prove that this does not happen for these curves.)
So there are three values of theta that are of interest: the value where the blue curve intersects the origin, the value where the blue and red curves intersect, and the value where the red curve intersects the origin. These three values form the limits of two integrals (no need for a double integral!) which are to be added to find the final area.
- Regardless of the coordinate system, the principle of finding the limits of integration is the same: find the minimum and maximum value of the independent coordinate that enclose the region of interest.
- In polar coordinates, for these two particular curves, the region of interest is best described as two parts. As theta sweeps anticlockwise, first it covers the area between the origin and the blue curve, until reaching the (non-origin) point of intersection between the blue and red curves. Then it covers the area between the origin and the red curve, up until the red curve meets the origin again.
- (This description would need to be modified for different curves. In particular, if the line segment between the origin and a point on either curve ever intersected the curve at a third point, you would need to take that into account. You can prove that this does not happen for these curves.)
- So there are three values of theta that are of interest: the value where the blue curve intersects the origin, the value where the blue and red curves intersect, and the value where the red curve intersects the origin. These three values form the limits of two integrals (no need for a double integral!) which are to be added to find the final area.
- To find those values of theta, simply use the polar-coordinate equations you have for the curves, and solve for theta when r is 0 (to find the intersections with the origin) and when the two curves have the same r (to find the curves' intersection with each other).
#1: Initial revision
by
r~~
·
2021-08-24T16:17:04Z (over 3 years ago)
Regardless of the coordinate system, the principle of finding the limits of integration is the same: find the minimum and maximum value of the independent coordinate that enclose the region of interest. In polar coordinates, for these two particular curves, the region of interest is best described as two parts. As theta sweeps anticlockwise, first it covers the area between the origin and the blue curve, until reaching the (non-origin) point of intersection between the blue and red curves. Then it covers the area between the origin and the red curve, up until the red curve meets the origin again. (This description would need to be modified for different curves. In particular, if the line segment between the origin and a point on either curve ever intersected the curve at a third point, you would need to take that into account. You can prove that this does not happen for these curves.) So there are three values of theta that are of interest: the value where the blue curve intersects the origin, the value where the blue and red curves intersect, and the value where the red curve intersects the origin. These three values form the limits of two integrals (no need for a double integral!) which are to be added to find the final area.