TRASNCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT (e). Given, 4x4gy = 36.first we find the areaof regionA=_JS₂A=> A= $1>) A =-3-²/314x²So [y]dx1² [9] 24/ 14x²=> A = ²2/²/²/ [x 14x² +9 sin³¹ (3) ] ² = 3 [957 +25] => [A = 6mNow, moment of inestia about x-axis!-y ² dydxIx = √√y²dA =>olydxn3S² S-3[³] =(²²-x²²x3Ix = (1²3²) ²³.2. & [√4x² (3x²³_45x) + 243 sin" ( 3 ) )].Tx =x293. = 6 TT² 1/2Stx dx= 6 π = ) | Ix=A]moment of Inertia about yaxis!-Ty = √√x²dA =)3So farex² dy dx= 2 [² x ²-1/²dx31/31F-X23==—= √²³²x²-19-x² dx3Ty = 2/1/12 - 21 [ 17x² (2x³²9x) +81 sin¹ (3)]27 7 (617)=) Tg = 38/²8 (81-17) = TTy==>) Ty = 2/A

The given curve isFirst, we find the area of the region

(e). Gliven, 4x4ay²=36. first we find the area of region A = $dA ²5*² => A= >> A = [₁ [y] => A= dlyde So [de= dx -4/314x2 -3 2 Ijds ${ [²√7x dx. ²/²/²/² [x14x² + 9 sin" (4) ] ² = 3 [9+ 2] =>(A = 6T / 3

(e). Gliven, 4x4ay²=36. first we find the area of region A = $dA ²5*² => A= >> A = [₁ [y] => A= dlyde So [de= dx -4/314x2 -3 2 Ijds ${ [²√7x dx. ²/²/²/² [x14x² + 9 sin" (4) ] ² = 3 [9+ 2] =>(A = 6T / 3

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter6: Applications Of The Derivative

Section6.CR: Chapter 6 Review

Problem 20CR

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TRASNCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT

(e). Given, 4x4gy = 36.
first we find the area
of region
A=_JS₂A
=> A= $1
>) A =
-3-²/314x²
So [y]dx
1² [9] 24/ 14x²
=> A = ²2/²/²/ [x 14x² +9 sin³¹ (3) ] ² = 3 [957 +25] => [A = 6m
Now, moment of inestia about x-axis!-
y ² dydx
Ix = √√y²dA =>
olydx
n
3
S² S
-3
[³] =(²²-x²²x
3
Ix = (1²3²) ²³.2. & [√4x² (3x²³_45x) + 243 sin" ( 3 ) )].
Tx =
x
293. = 6 TT
² 1/2
Stx dx
= 6 π = ) | Ix=A]
moment of Inertia about yaxis!-
Ty = √√x²dA =)
3
So fare
x² dy dx
= 2 [² x ²-1/²
dx
31/31F-X2
3
==—= √²³²x²-19-x² dx
3
Ty = 2/1/12 - 21 [ 17x² (2x³²9x) +81 sin¹ (3)]
27 7 (617)
=) Tg = 38/²8 (81-17) = T
Ty
=
=>) Ty = 2/A

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