Double integral approach to centroid location: ...... again addressing x- , for example.
Here we may choose instead to integrate using an elemental area dA = dxdy, i.e.,
where each dimension is infinitesimal. Now our expression is going to have to be
integrated with respect to both variables, holding one constant and integrating over the
other, and then integrating over the second. So for the triangle:
y = m' x
- (x , y)
centroid is
somewhere here
b
⌠
⌡ x dA
x-
⌠
⌡⌠
⌡ x dx dy
= ⌠ dA
⌡
=
A
If we integrate 1st on x , then on y, so y is fixed to begin with and we find that, for any
arbitrary area dxdy, x runs from the left hand value of
y
(we can't just put in zero, because then we've eliminated y already!!) to b and then we
m'
repeat the integration as y goes from 0 to m'b (that we'll call "h").
h
⌠
⌡
=
0
[
b
⌠ x dx
⌡
h
] dy
y/m'
=
1
2 bh
y
h
2
⌠ b2 - ( ) dy
⌡
m'
=
0
bh
h
and m' = b
x2
b
⌠
⌡ 2 |y/m' dy
=
0
1
2 bh
1 h3
b2h - m'2 3
so x- =
bh
h
b2h - b2 3
bh
=
2
3 b
2
and y- is also going to be 3 h down from the corner towards the right angle just like x2
h
is 3 b from the corner towards the right angle, so y- = 3
Locate the centroid of the paraboloid of revolution, generated by rotating the
shaded area about the y-axis.
……… by the method of multiple (here, triple) integrations
⌠
⌡ y dV
⌠
⌡⌠
⌡⌠
⌡ y dx dy dz
V
y= ⌠ dV
so
y= ⌠ ⌠ ⌠ dx dy dz
⌡
⌡⌡⌡
V
and the questions are ... in what order .... and what to use as limits? Let's just consider
the numerator first, ⌠
⌡⌠
⌡⌠
⌡ y dx dy dz
Well, seeing we want to finish up with the value of y, let's do that one last, ..........
and let's begin with x (why not?!). So the expression becomes:
⌠
⌡ y
[ ⌠⌡ [⌠⌡
dx
] dz ]dy
and what as limits? Well, since this is symmetrical, then z = x, so the limits for the inner
integration are from 0 to z.
Then, for the center integration, z goes from 0 to 10 y , since z2 = 100y.
And for the outer integration, y goes from 0 to 100. So the expression is now:
=
100
⌠
⌡ y
0
[ 10⌠⌡ y [⌠⌡z
0
0
dx
] dz ]dy
=
=
100
⌠ y
⌡
0
100
⌠ y
⌡
[ 10⌡⌠ y [x |z0 ] dz ]dy
[
0
=
0
10 y
⌠
⌡
]
z dz dy =
100 -
⌠y
⌡
0
100
⌠
⌡
100y
2
y
[
0
dy
0
Now, putting back the denominator, to calculate y100
⌠ y
⌡
y-
=
100y
dy
2
0
100
100y
⌠
dy
⌡
2
[ 50y3 |1000 ]
[25 y2 |1000 ]
3
=
0
=
50 106
3
25 104
= 66.7mm
z2
2
|10√y
] dy
0