WEEK 16

This week we learned how to find volumes of disks and of washers. Let me review how to find area first.

To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.

Example:
Question: Find the area of the region enclosed by y= -7x^2 + 14x + 152 and y = -4x^2 -x + 2 between x=0 and x=8.

Solution: You first graph both functions and see that the first equation given is on top.
So you set up your equation as 8(int)0 (-7x^2 + 14x + 152) - (-4x^2 -x + 2).
You then distribute the negative and combine like terms to get: 8(int)0 (-3x^2 + 15x +150).
You then intergrate and get: -x^3 + (15/2)(x)^2 + 150x 80. You then plug in f(8) - f(0) to get 1168 units squared.

So now on to volume.
First of all: Disks are solid objects with one equation bounded by the equation and usually an axis. Washers are objects bounded by two equations.

Formula for Disk:

(Pi) b(int)a [R(x)]^2.

Formula for Washers:

(Pi) b(int)a (top eq.)^2 - (bottom eq.)^2 dx.


Something I do not get from this week is how to solve equations when you have to shift.

An Example would be number 13 from the 7.2 homework worksheet.

the directions say to find the volumes of the regions bounded by both graphs and the lines given.

y = x^2 , y =4x - x^2

a) the x-axis
b) the y-axis

Overall, I think I am starting to get the process of working these types of problems a little better, but just need a little extra practice and maybe somebody to re-explain the shifting problems.