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Chapter 6: Q. 51 (page 523)

The region between the graph of f(x)=(x-3)2+2and the x-axis on [0, 6], revolved around the y-axis

Short Answer

Expert verified

The volume is180蟺cubicunits

Step by step solution

01

Given information

We are given a function asf(x)=(x-3)2+2

02

Find the integral and evaluate it

We know that integral can be given as V=2蟺鈭�abr(x)h(x)dx

The axis of revolution is y-axis hence the radius is r(x)=x

and the height can be given as h(x)=(x-3)2+x

Substituting the values in the formula we get,

role="math" localid="1650733726902" V=2蟺鈭�06x(x2-6x+11)dxV=2蟺鈭�06(x3-6x2+11x)dxV=2蟺[x44-2x3+112x2]60V=180蟺

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