This week we started learning about revolving disks and washers. We just took a quiz today on this subject, and it was a bit challenging but I think I did alright. I think the problems I struggle with the most are the ones that involve rotating a shape around a line that isn't an axis. For example, there was one on the quiz that was rotating around an y= line and it had a hole. It was hard to figure out how to find the larger radius and the smaller radius. This is because you end up having to subtract equations from other equations and figure out which ones are which to be able to make the correct integral. I have realized that making the rectangle is the best way to visualize what you are trying to do and the shape that the graphs are trying to make. It also helps you to see the right way to write the integral and how the integral will look.
I have also realized that it is much easier to separate the problems into chunks because you can figure out the integral much quicker that way. First, I always figure out whether the integral is going to be in terms of x (dx) or y (dy). This then allows me to realize whether the equations I am using are going to be in terms of x or in terms of y. (I also always try to make sure when I plug the equations into the calculator, they are always in terms of x). Then, if the graph does not have a hole in it to make it a washer, I know that the integral will most likely be just that equation squared. But if there is a hole between the equation(s) and the line you are rotating it over, I know that I have to split the next step into two parts. These two parts include finding the bigger radius (distance between the top line and the line you are rotating it over) and the smaller radius. First, you must find the bigger radius because that is what is going to come first in the integral. This involves taking whatever equations you are given and the line that it is being rotated around, and finding the greatest distance. Then, you find the smaller radius, which usually is just the height of the rectangle itself.
As I have said before, it is nice to see how everything in calculus seems to relate to everything else. What really helped me with this particular section was knowing how to find the areas between curves. Without knowing how to do that, it would have been very difficult to figure out some of the problems with washers. It would also be difficult finding the integral itself because of the trying to find what variable it is supposed to be in terms of.
I have also realized that it is much easier to separate the problems into chunks because you can figure out the integral much quicker that way. First, I always figure out whether the integral is going to be in terms of x (dx) or y (dy). This then allows me to realize whether the equations I am using are going to be in terms of x or in terms of y. (I also always try to make sure when I plug the equations into the calculator, they are always in terms of x). Then, if the graph does not have a hole in it to make it a washer, I know that the integral will most likely be just that equation squared. But if there is a hole between the equation(s) and the line you are rotating it over, I know that I have to split the next step into two parts. These two parts include finding the bigger radius (distance between the top line and the line you are rotating it over) and the smaller radius. First, you must find the bigger radius because that is what is going to come first in the integral. This involves taking whatever equations you are given and the line that it is being rotated around, and finding the greatest distance. Then, you find the smaller radius, which usually is just the height of the rectangle itself.
As I have said before, it is nice to see how everything in calculus seems to relate to everything else. What really helped me with this particular section was knowing how to find the areas between curves. Without knowing how to do that, it would have been very difficult to figure out some of the problems with washers. It would also be difficult finding the integral itself because of the trying to find what variable it is supposed to be in terms of.