Show that the area of the shaded region equals 1/ 6 for all values of z .

Name: Problem 81, Chapter 5: Calculus
Uploaded: 2020-11-24T14:31:08-08:00
Duration: 4 min 52 s
Channel: Linda Hand
Description: Show that the area of the shaded region equals 1/ 6 for all values of z .

Show that the area of the shaded region equals 1/ 6 for all...

Key Concepts

Area Calculation

Area calculation involves determining the size of a region by identifying its boundaries and applying relevant mathematical techniques such as geometric formulas or integration. This concept is crucial for problems involving shaded regions or areas bounded by curves, where precise computation is necessary to arrive at an exact value.

Invariant Properties

Invariant properties are features of a geometrical configuration that remain constant even when certain parameters change. In the context of the problem, showing that the area of the shaded region remains equal to 1/6 for all values of the parameter highlights an important invariant feature, meaning the area does not depend on the parameter z.

Parametric Analysis

Parametric analysis examines how the behavior or value of an expression depends on one or more parameters. In this problem, despite the presence of a variable z, the analysis demonstrates that the outcome—specifically, the area of the shaded region—remains fixed at 1/6, indicating a particular structure or symmetry in the mathematical relationship involved.

Transcript

00:01 We want to find the area between 1 minus x squared in this line that goes from z to z plus 1.

00:08 Okay, and what's going to happen is in the end we're going to get one sixth.

00:12 So it's going to say no matter what z you use, you're going to get one six for the answer.

00:17 Okay, and it's not just random.

00:19 Okay, this special, it's a special curve that makes it work this way.

00:23 Okay, so the first thing i need is the equation of the line, which i've already worked out for you.

00:27 I found the two points, 0 .1, 0 .2, plugged them into the slope formula.

00:34 And that's when i noticed that i have z plus 1 minus z on the bottom, which turns into 1.

00:40 So that gave me a hint that i need to keep this in mind.

00:45 So the slope turns out to be z squared minus z plus 1 squared, which i factored, and i changed it to this.

00:54 And then i wrote the equation of the line.

00:56 Okay, now leave it all.

00:57 Factored up like that because that's the trick to working this problem.

01:02 I've worked it several times.

01:03 So i know.

01:04 Okay, so the area is from z to z plus one.

01:09 Top and the top is the parabola, 1 minus x squared, minus the bottom, which is the line, x minus z plus 1 minus z squared, dx for the whole thing.

01:35 Okay, so i'm going to work on that inside the parentheses because i i've tried to integrate it this way.

01:40 It just makes a big of mess.

01:42 Okay, so one subtracted off.

01:45 Okay, so i'm just going to work down here.

01:47 So i notice i have a minus x squared and a minus z squared.

01:51 So i have z squared minus x squared.

01:55 And then i have plus 2z plus one times x minus z.

02:03 So i decide i'm going to change the order here...