The problem:
Please rotate y=x+4, y=x^2+2 around the line y=6.
The graph:
Please rotate y=x+4, y=x^2+2 around the line y=6.
The graph:
The first step is to figure out what the integral will be in terms of. For this problem, since it is being rotated around y=6, the integral will be in terms of x (according to the rectangle). Then you find the bounds. For this problem, the bounds will be the intersections of the two equations, which is -1 to 2. The next step is to figure out what the height of the rectangle is. This requires you to find the big radius and the small radius. In this case, the big radius will be the distance from the line of rotation, y=6 to the bottom of the curve y=x^2+2. This means that the first part of the integral will be
(6-(x^2))^2. The whole thing is squared because that is a part of the volume equation. Then we must find the smaller radius, which is where I got caught up. The smaller radius is essentially the same thing, except it is from the line y=6 to the cure y=x+4. In the quiz, I got this mixed up and said that it was from the line y-x=4 to the cure y=x^2+2. This is not the small radius because it doesn't include what you want it to which is the empty space or the "hole". When you figure all of this out, the integral should look like this:
pi x the integral from -1 to 2 of (6-(x^2))^2 - (6-(x+4))^2 dx.
Here is another problem:
Rotate y=x and y=x^2-4x+4 around y =4.
The graph:
(6-(x^2))^2. The whole thing is squared because that is a part of the volume equation. Then we must find the smaller radius, which is where I got caught up. The smaller radius is essentially the same thing, except it is from the line y=6 to the cure y=x+4. In the quiz, I got this mixed up and said that it was from the line y-x=4 to the cure y=x^2+2. This is not the small radius because it doesn't include what you want it to which is the empty space or the "hole". When you figure all of this out, the integral should look like this:
pi x the integral from -1 to 2 of (6-(x^2))^2 - (6-(x+4))^2 dx.
Here is another problem:
Rotate y=x and y=x^2-4x+4 around y =4.
The graph:
This is essentially the same problem. The bounds this time are 1 and 4 because those are the intersections. The bigger radius is (4-(x^2-4x+4))^2 and the smaller radius is (4-(x))^2. So the integral will look like this:
pi x the integral from 1 to 4 of (4-(x^2-4x+4))^2 - (4-(x))^2 dx.
pi x the integral from 1 to 4 of (4-(x^2-4x+4))^2 - (4-(x))^2 dx.