Blog Post #9

1.) Identify the difference between rotating around a vertical line and a horizontal line. When rotated around a vertical line, the washers are stacked vertically, one on top of another. When rotated around a horizontal line, the washers are stacked horizontally, side by side. Which variable goes with each one? When rotated around a vertical line, you will be working in terms of y. When rotated around a horizontal line, you will be working in terms of x.

2.) What is the difference between creating a washer and a disk? When a washer is created, there is a hole within the figure, resulting in subtraction. When a disc is created, there is no subtraction. A washer is created when the shaded region of the figure does not connect with the line that the function is being rotated around. A disc is created when the shaded region of the line does connect with the line that the function is being rotated around.

3.) Identify two examples that use rotating about a vertical and horizontal line.

Example 1: Rotated about a horizontal line.

X=0, Y=X^2, Y=4, Rotated around Y=6

Because it is being rotated around a horizontal line, it is in terms of X. We can use X=0 and Y= 4 to determine our interval. When we plug 4 in for Y in Y=X^2, we get X=2. We already know that X=0, so we know the interval is from 0 to 2. We know that x^2 is our radius because it is what changes. We are required to subtract our radius from six because we need to exclude the portions between 0 and 2 that are not filled in. We also need to subtract 6 from 4 because there is a a gap on that interval. We will also pull out pi.  It will look like pi(0 to 2) ((6-x^2)^2)-(6-4)^2. When you simplify it will be pi(36-12x^2+x^4-4 dx). When you take the anti derivative, you get pi(32x-4x^3+1/5x^5). When you plug in the interval 2 (You do not need to plug in zero as it would be subtracted from 2 and zero will give you zero) you should get pi192/5.

Example 2: Rotated about a vertical line.
X=0, X=y-y^2, revolved around the y-axis.
To find our interval, we must set y-y^2 equal to zero. When we do, we get 0=y(1-y) which factors to give us y=0 and y=1. Our interval is 0 to 1. Because it revolves around a vertical line, it will be in terms of y. Our radius will be y-y^2 because it is what changes. The setup will be pi(0 to 1) ((y-y^2)^2)dy. The anti derivative is pi(0 to 1) y^3-1/2y^4+1/5y^5. All we have to do is plug in one because the simplification of zero will be subtracted from the simplification of one. When one is plugged in it should be pi/30.