00:04
We want to represent the area of the region between these two curves by using a definite integral.
00:12
So let's take a look at the graphs of these two curves.
00:16
So they both represent circles, circles with different centers.
00:22
But the area, the region that we are speaking about is this area right in here.
00:32
Okay, the area between the function, so we're looking at this little kind of football -shaped, kind of almost oval -shaped region in here.
00:42
This is the area.
00:43
This is the region that we want to find the area of.
00:49
Now we have two choices.
00:51
We can integrate with respect to the x -axis, which wouldn't be too bad because the blue function stays above the green function.
01:01
You know, from x is zero to x is two.
01:06
And the same thing would actually happen if we integrated with respect to the y -axis, the blue function is staying above the green function.
01:15
So it doesn't matter which variable x or y we integrate with respect to.
01:22
So let's just integrate with respect to the x -axis.
01:25
And that means we have to rewrite both of these equations where we, y is written as a function of x.
01:34
So we have to take each of these equations, algebraically manipulate them, so y is written as a function of x.
01:46
All right, so if we start with this one over here, subtracting x squared from both sides, we get y squared equals 4 minus x squared, and then y would equal to positive or the negative.
02:04
We're actually only going to end up needing one of those.
02:07
But y would equal to positive or the negative square root of 4 minus x squared.
02:16
Let's take a look real quick.
02:20
Y equals the square root of 4 minus x squared.
02:29
Okay.
02:30
All right.
02:30
See it in purple? now it's blue.
02:33
Now it's purple.
02:34
Okay.
02:35
So you can see this is one of the functions that we need.
02:38
So we don't need the whole blue circle anymore.
02:40
So let's get rid of that.
02:43
Now we're going to do the same thing with the green circle equation.
02:47
Okay, we're going to rewrite this.
02:49
So, y is a function of x.
02:52
And we'll either get the top half of the circle or the bottom half.
02:57
We're going to want the bottom half.
03:01
All right.
03:01
So now i'm manipulating this.
03:04
Y minus 2 squared equals 4 subtract, the x minus 2 squared.
03:19
We've basically just subtracted the x minus 2 squared from both sides of the equation.
03:25
Now we've got to take the square root of both sides.
03:35
Doing so, the square root cancels out the second power.
03:38
We get y minus 2 equals the positive or the negative of the square root of 4 minus x minus 2 squared.
03:54
I'll show you in a minute.
03:55
We're obviously only going to need one of these.
03:58
Want to end up being the negative one.
04:01
What gives it away is we want the bottom half of the green circle.
04:05
So that's why we're going to use the negative one as opposed to the positive square root.
04:09
We're going to want the negative square root.
04:11
But we'll show you.
04:12
We'll see that in a minute.
04:14
Let's finish solving this for y.
04:16
Just add the two to both sides...