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Chapter 6: Q. 52 (page 523)

The region between the graph of f(x)=x2and the x-axis on [0, 3], revolved around the line y = 10

Short Answer

Expert verified

The volume is165.5蟺cubicunits

Step by step solution

01

Given information

We are given a functionf(x)=x2

02

Find the integral and evaluate it

The volume of the integral can be given as V=2蟺鈭�cdr(y)h(y)dy. The axis of revolution is y = 10

Hence the radius can be given as r(y)=10-y

And the height of the function can begiven as h(y)=y

Substituting the values in the function we get,

role="math" localid="1650734209457" V=2蟺鈭�09(10-y)(y)dyV=2蟺鈭�09(10y-yy)dyV=2蟺[203y32-25y52]90V=2蟺[82.8]V=165.5蟺

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Most popular questions from this chapter

Consider the region between f(x)=xand the x-axis on [0,4]. For each line of rotation given in Exercises 27鈥�30, use four disks or washers based on the given rectangles to approximate the volume of the resulting solid.

We have been given

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

f(x)=13x3/2-x1/2,a,b=0,1

Consider the region between the graph of f(x)=3xand the x-axis on [1,3]. For each line of rotation given in Exercises 31鈥� 34, use definite integrals to find the volume of the resulting solid.

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

f(x)=(2x+3)3/2,a,b=-1,1
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