Determine the area of Ωand locate the centroid.

Name: Problem 37, Chapter 10: Calculus: One and Several Variables
Uploaded: 2021-11-08T05:38:42-08:00
Duration: 6 min 57 s
Channel: Harmender Singh Yadav
Description: Determine the area of Ωand locate the centroid.

Key Concepts

Moment Integrals

Moment integrals, specifically the first moments, involve integrating the coordinates multiplied by the density (or simply the coordinate values in the case of uniform density) over a region. These integrals are used to determine the location of the centroid by quantifying how the area is distributed with respect to the coordinate axes.

Area of a Plane Region

This concept involves calculating the area enclosed by a closed curve or region in the plane. In many cases, the area is determined by setting up and evaluating a double integral over the region, taking into account any boundaries or limits defined by the region.

Centroid of a Region

The centroid, often regarded as the center of mass for a uniformly dense region, is the point at which the region would balance perfectly. Its coordinates are computed using the first moments of the region (integrals of x and y over the area) divided by the total area.

Double Integration

Double integration is the process of integrating a function in two variables over a specified region in the plane. It is the primary tool for computing areas as well as the moments needed for centroid calculations.

Transcript

00:01 In this question, an equation of a parabola is given as x squared equal to 4 c y and omega is the region between the latest rectum and the parabola.

00:10 So we need to find out the area of this region and also the centroid of this region.

00:20 So first of all, we will plot the graph of this parabola and shade that region of which we need to find out the area and the centroid.

00:27 So we plotted the graph and also shaded the region of which we need to find out the area.

00:33 So first of all we will calculate this area as this is the area under the line which is the latest rectum which can be stated as y equals to c.

00:45 And the area between this line y equals to c and the parabola that is y equals to x square over 4c can be calculated as the area equal to integral for the limit minus 2c to 2c as this area is extending from x equals to minus 2c to 2c and for y dot d x and y will be equal to for for this line will be c and minus for this parabola it will be x square over 4c so x square over 4c dot d x now we will integrate this and we will find out this area so this area will be equal to that is the integral of for the limit as we know that this is a even function so we can take this limit from 0 to 2c and by multiplying this integral by 2 so we will get it as c minus x square over 4c for the limit 0 to 2c and here d x now we will integrate this and we will get this area as twice of integral of c will be c will be c x minus integral of x square over 4c will be x cube over 3 times 4 c that will be 12c so for the limit 0 to 2c executing this limit we will get this area as that is 8 over 3 c square so this area will be equal to 8 over c so 8 over 3 c square square units...

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