(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

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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(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
Transcribed Image Text:(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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