Post #16

Last week in calculus, we learned how to find volume by disk and how to use washers.

Disks is only used with solid objects and the formula is pi interval a, b [R(x)]^2 dx
An example would be: Evaluate the integral that gives the volume of the SOLID formed by revolving the region about the x-axis. y=-x+1
The first step would be to graph the equation which I can't show on here
By graphing it, it will show what interval you are looking at. In this case it is [0,1]
Then plug into your equation: pi integral of (0,1) [-x+1]^2
Next you have to square the equation in order to integrate it: x^2-2x+1
Integrate: 1/3 x^3-2(1/2x^2) +x which simplifies to 1/3x^3 - x^2 +x
Then all you have to do is plug in your intervals: 1/3(1)^3-(1)^2+1-[0] = 1/3 pi.

Washers is used when there is two equations.
The formula is pi integral of (a,b) top^2 - bottom ^2 dx
Example: Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines.
y=6-2x-x^2 , y=x+6 about the x-axis
First step just as in disks is to graph the equations.
Nest you set your equations equal to each other to find the interval you are looking at.
6-2x-x^2=x+6
-3x-x^2 = 0
solve for x: -x(3+x)
x=0 and x=-3 therefore your interval is (-3, 0), 0 on the top because the bigger number is b.
After you find your interval, you have to use your graph to determine which equation is on top. If unsure, you can always plug into your calculator if it is allowed.
In this problem, 6-2x-x^2 is on top so you subtract x+6 from it
(6-2x-x^2)^2 - [x+6] ^2
Then you would have to square each and add like terms
Remember (6-2x-x^2)^2 is solved by multiplying (6-2x-x^2)(6-2x-x^2)
Then you would integrate in and plug your intervals into the integral.
Your answer should be 243/5 pi. Don't forget the pi.

(b) same equations about the line y=3
It's the same steps except you would have to subtract 3 from each equation
6-2x-x^2-3 - [x+6-3]
simplifies to: (3-2x-x^2)^2 - [x+3]^2
The interval would remain the same so on (-3, 0)
After following the same steps as problem a which were square everything, simplify, integrate, then plug in, your answer should be 108/5 pi.

I am having trouble with trig inverse integration, which I think we also learned last week. I have no idea where to start or the what to do at all so if anyone can explain it to me from start to finish I would greatly appreciate it.