/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 73 Prove that the centroid of a cir... [FREE SOLUTION] | 91影视

91影视

Prove that the centroid of a circle is the center of the circle.

Short Answer

Expert verified

the centroid of a circle is the center of the circle.

The centroid of a circlex2+y2=a2is at(x,y)=(0,0)and the center is also at(0,0).

Step by step solution

01

Step 1. Given information.  

The given statement is the centroid of a circle is the center of the circle.

02

Step 2. x coordinate of the centroid

Consider a circle x2+y2=a2of constant density with a center at (0,0).

xcoordinate of the centroid of the circle is following.

x=x(x,y)dxdy(x,y)dxdyx=x=aay=a2x2a2x2xdxdyx=aay=a2x2a2x2dxdy

change the system into a polar system by substituting x=rcos,y=rsin&dxdy=rdrd.

localid="1650439208659" x==02=0a(rcos)rdrd=02r=0ardrdx=r=0ar2dr=02cosdr=0ar2dr=02dx=(a33)(0)(a33)(2)x=0

03

Step 3. y coordinate of the centroid

ycoordinate of the centroid of the circle is following.

y=y(x,y)dxdy(x,y)dxdyy=x=aay=a2x2a2x2ydxdyx=aay=a2x2a2x2dxdy

change the system into a polar system by substitutingx=rcos,y=rsin&dxdy=rdrd

y==02r=0a(rsin)rdrd=02r=0ardrdy=r=0ar2dr=02sindr=0ar2dr=02dy=(a33)(0)(a33)(2)y=0

the centroid of the circle is p(x,y)=(0,0).

So the centroid of a circle is the center of the circle.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.