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91Ó°ÊÓ

The region between the graph of f(x)=9-x2and the x-axis on [0, 3], revolved around the x-axis

Short Answer

Expert verified

The volume is23.8Ï€cubicunits

Step by step solution

01

Given information

We are given a functionf(x)=9-x2

02

Find the integral and evaluate it

We know that integral can be given as V=2π∫cdr(y)h(y)dy

As the axis of revolution is x-axis we have r(y)=y

and h(y)=9-y

Substituting the value in the integral as

V=2π∫03y9-ydyV=2π[11.9]V=23.8π

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