Find b such that the line y=b divides the region bounded by the graphs of the two equations into

Find b such that the line y=b divides the region bounded by...

Key Concepts

Definite Integration

Definite integration is the method used to calculate the exact area under a curve between two set points. This concept is central for determining the total area of a region bounded by one or more curves.

Area Between Curves

This concept involves computing the area enclosed between two functions by integrating the difference of the functions over an interval established by their intersection points. It is a foundational technique in finding the region's total area.

Equal Partitioning of Areas

Partitioning a region into two areas of equal size requires setting up an equation where one integral, representing the area on one side of a dividing line, equals half of the total area. This concept involves solving for a variable that defines the partition boundary.

Symmetry in Functions

When a region exhibits symmetry with respect to an axis, the problem can often be simplified by computing the area over a symmetric interval and then doubling the result. Recognizing symmetry helps reduce the complexity of the integration process.

Transcript

00:01 The first thing i need to do in this problem is figure out what the area is of this equation, y equals 9 minus x squared.

00:13 I'm assuming most math students are good with, you know, it's a downwards problem for the negative x squared, shifted up 9.

00:22 And what might actually be a good idea is to rewrite this in terms of x, you know, so may add x squared to the left side is equal to 9 minus y and then square root now i'll actually do this in red because uh what i'm thinking about is this is an even function so if i just do the positive version i can just get this but if i'm splitting this in half and i'm i guess what my goal is is to find the horizontal line where and it's not a very good horizontal line i don't know exactly where it is because i haven't solved this yet, but where this area is equal to this area, because if that's the case, then all of this area will equal all of that.

01:13 Yeah, i guess that's my ultimate goal.

01:18 So what i would do is figure out the integral from 0 to 9 of this function, 9 minus y, d .y, and as i do the anti -derivitum, it might actually be beneficial if you think about this as to the one -half power, because what you can do, 9 minus y, add 1 to the exponent.

01:47 So 1 -half, add 1 is 3 halves, but then you have to multiply by the reciprocal.

01:51 But don't forget about the chain rule.

01:53 If you took the derivative of this, you would have a negative in front.

01:56 So we also need a negative in front.

01:59 And that's from 0 to 9.

02:01 So what's nice about this is as you plug in your bounds, like nine in here, nine minus nine is zero to any power still zero.

02:12 So we're looking at zero minus, you know, that negative two -thirds times plugging in zero in here.

02:22 Nine minus zero is nine to the three -half power.

02:26 So remember this is the square root of nine, which is three cubed is 27...

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