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Calculus II, Section 8.3, #28
Applications to Physics and Engineering
Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the
exact coordinates of the centroid.1
y = sin (x),
y = 0,
0≤x≤π
Here’s a sketch of the region:
y
1
x
π
2
π
From the symmetry about the line x = π2 , it seems that the x-coordinate of the centroid will be
visually estimate the y-coordinate to be ≈ 0.35.
π
2.
I’d
The relevant formulas for the centroid (center of mass of the region) are
Z
Z
o
1 b
1 b1n
2
2
dx
[f (x)] − [g(x)]
x̄ =
x [f (x) − g(x)] dx,
ȳ =
A a
A a 2
For this problem, a = 0, b = π, f (x) = sin(x), and g(x) = 0.
The area, A, is given by
Z π
A=
sin (x) − 0 dx
0
π
= [− cos (x)]0
= − cos (π) − − cos (0)
= − (−1) − − (1)
=2
So
x̄ =
=
=
=
=
=
Z
1 π
x [sin (x) − 0] dx
2 0
Z
1 π
x sin (x) dx
2 0
1
[−x cos (x) + sin (x)]π0
2
1
[(−π · cos (π) + sin (π)) − (−0 · cos (0) + sin (0))]
2
1
[π]
2
π
2
and
1
ȳ =
2
1 Stewart,
Z
0
π
1 2
sin (x) − 02 dx
2
Calculus, Early Transcendentals, p. 567, #28.
Integration by parts:
Z
x sin (x) dx
Let u = x, so dv = sin (x)dx.
Then du = dx and v = − cos (x).
Z
= x · − cos (x) − − cos (x) dx
Z
= −x cos (x) + cos (x) dx
= −x cos (x) + sin (x)
Calculus II
Applications to Physics and Engineering
1
=
2
Z
0
π
1
sin2 (x) dx
2
Using the trigonometric identity sin2 (x) =
1−cos (2x)
2
=
1
2
(1 − cos (2x)),
Z
1 π1 1
=
· (1 − cos (2x)) dx
2 0 2 2
Z π
1
=
1 − cos (2x) dx
8 0
π
sin (2x)
1
x−
=
8
2
0
1
sin (2π)
sin (0)
=
π−
− 0−
8
2
2
1
= [π − 0 − 0 + 0]
8
π
=
8
= 0.3927
Thus, the exact coordinates of the centroid are
π π
2,8
or about (1.5708,0.3927).
y
1
b
x
π
2
π