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91Ó°ÊÓ

Use four slices to construct an approximation of each of the

quantities described in Exercises 25–30. In each case include

a labeled diagram and an explicit list of the values of ∆xand

each xkand xk*(or ∆yand each ykand yk*).

The mass of a cylindrical rod with radius of 2 centimeters and length of 24 centimeters whose density at a point x centimeters from the left end is ÒÏ(x) = 10.5 + 0.01527x2 grams per cubic centimeter.

Short Answer

Expert verified

ky=2.35ftkx=1.8ft

Step by step solution

01

Step1. Given Information

radius=2cm

length=24cm

density at left end =ÒÏ(x)=10.5+0.01527x2gmcm3

02

Step2. Calculate

dm=ÒÏdv=ÒÏ(Ï€x2)dydm=ÒÏÏ€1y2dydIy=12dmx2

03

Step3. Continue

dIy=12(ÒÏÏ€1y2dy)(1y2)=12ÒÏÏ€dyy4∫dIy=12ÒÏπ∫0.2541y4dy=Iy=134+ÒÏ

04

Step4. Continue

c=12[ÒÏÏ€(42)(0.25)](42)=32Ï€ÒÏ∫dm=m=ÒÏπ∫0.2541y2dy⇒m=24.34ÒÏ

05

Step5. Continue

kx=Ixm,ky=Iym

06

Step6. Continue

ky=134ÒÏ94.34ÒÏ⇒ky=2.35ft.

07

Step7. Continue

dI'x=14dmx2+dmy2=14(ÒÏÏ€dyy)1y2+(ÒÏÏ€dyy)y2=ÒÏÏ€(14y4+1)dy∫dI'x=∫0.254ÒÏÏ€(14y2+1)dy

08

Step8. Continue

I'x=(-112y3+y)|0.254⇒I'X=28.23y

09

Step9. Continue

I''x=14[ÒÏÏ€(42)(0.25)](42)+[ÒÏÏ€(42)(0.25)](0.1252)I"x=50.46ÒÏ

10

Step10. Continue

Ix=I'x+I"x=28.53ÒÏ+50.46ÒÏIX=78.99ÒÏkx=78.99ÒÏ24.37ÒÏ⇒kx=1.8ft.

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