Use provided information please ShapeTriangular areaQuarter-circularareaSemicircular areaQuarter-ellipticalareaSemiellipticalareaSemiparabolicareaParabolic areaParabolic spandrelGeneral spandrelCircular sectorShapeQuarter-circulararcSemicircular areArc of circleIZI10월가을에43L|---0HIFy=kx2XC•y=kx"|——I.C₁ța0|X.oa013hty!CDA4bTx03π03a0n+n+22r sin a3aX2r0r sin aay454b3π4b3π3153h10n+1h4n+20yak2/1/k0AreabhT²Rab4Kab2ah4ah3ahahn+1TLength플MY2ar Use direct integration to find the centroid of the area shown in terms of a and h.y = kx²(a,h)X

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Physics

Use direct integration to find the centroid of the area shown in terms of a and h. y = kx² (a,h) X

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Introduction And Vectors

Section: Chapter Questions

Problem 32P

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Question

Use provided information please

Shape
Triangular area
Quarter-circular
area
Semicircular area
Quarter-elliptical
area
Semielliptical
area
Semiparabolic
area
Parabolic area
Parabolic spandrel
General spandrel
Circular sector
Shape
Quarter-circular
arc
Semicircular are
Arc of circle
IZI
10월가을에
43L
|---
0
HI
F
y=kx2
X
C•
y=kx"
|——I.
C₁
ța
0|
X.
oa
0
13
h
ty!
C
DA
4
b
T
x
0
3π
0
3a
0
n+
n+2
2r sin a
3a
X
2r
0
r sin a
a
y
45
4b
3π
4b
3π
315
3h
10
n+1h
4n+2
0
y
ak
2/1/
k
0
Area
bh
T²
Rab
4
Kab
2ah
4ah
3
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n+1
T
Length
플
MY
2ar

Use direct integration to find the centroid of the area shown in terms of a and h.
y = kx²
(a,h)
X

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