(b). Given y=8x³, y=0, x=1first we find the area!A = SS dALo fox³ dydx28x3=)A={ [y] oe' de = 8√' x³dx['A = 8. 12/²7/1[A=2Now, moment of inertia about x-axis!Ix = Sfa²dA =) ( 18x³y2dydx3 [19³] 8* dx = 50 (²x²+²x8x3x9dx110Ix= 313 [7²]yosuJ.you5123X10Ty = √√x²dA =)moment of inertia about y-axis!2[x². [yj³x³ deL'xz· 128 . (2) -) Tx²= 128-A=15JóI 18x³ xzdydz=8 5 x 5 xx==1Iy = 8.1 ²7 16=) Ty = // = ²/3 (2) =) | Ty= 3 A43

In this question, we will check the given answer is correct or not.

(b). Given y=8x²°, y=0, x = 1₁ first we find the area! 8x3 A = SS dA I fox³ dydx [ [y] 8x³² dx = 8 (²x³dx =) A = 1. A = 8. 12/²7/1 Ix= 312 [#²] J you => [A=2] Now, moment of inertia about x-axis! Ix = Say²da =) S fox³yzdydx 8x3 3 ( [9³] 8x dx = 510 (x1de x9dx - 5 512 Ty = √₁₂²dA =) 3X10 moment of inertia about y-axis! youx? = 128 (2) →) Tx= 128-A 8x3 Só joz xezdydz x=1 [²x², jyye²de = 8 ('xsde [y] 8x3 xSdx 6 Iy = 8.1 ²² 1² ⇒ & = 4 8 6

(b). Given y=8x²°, y=0, x = 1₁ first we find the area! 8x3 A = SS dA I fox³ dydx [ [y] 8x³² dx = 8 (²x³dx =) A = 1. A = 8. 12/²7/1 Ix= 312 [#²] J you => [A=2] Now, moment of inertia about x-axis! Ix = Say²da =) S fox³yzdydx 8x3 3 ( [9³] 8x dx = 510 (x1de x9dx - 5 512 Ty = √₁₂²dA =) 3X10 moment of inertia about y-axis! youx? = 128 (2) →) Tx= 128-A 8x3 Só joz xezdydz x=1 [²x², jyye²de = 8 ('xsde [y] 8x3 xSdx 6 Iy = 8.1 ²² 1² ⇒ & = 4 8 6

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter2: Graphical And Tabular Analysis

Section2.1: Tables And Trends

Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...

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