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تاريخ التسجيل : 21/09/2008
 |  موضوع: centroid method of integration  2008-09-26, 7:19 pm | |
| Centroid -
Method of Integration - 1  | | Example Problem
<table cellSpacing=0 cellPadding=5 align=right border=0><tr align=middle><td> </TD></TR>
<tr align=middle><td>Fig. 1</TD></TR></TABLE>
Use integration to locate the centroid of the shaded area shown in Fig. 1. |
 | | Solution
The centroid of an area is the location at which the entire area could be concentrated and it would have the same "moment" as the sum of the "moments" of the individual pieces of area. This sum is called the first moment of area.
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(1)</TD></TR></TABLE>
Finding the Area
<table cellSpacing=0 cellPadding=5 align=right border=0><tr align=middle><td> </TD></TR>
<tr align=middle><td>Fig. 2</TD></TR></TABLE>
To find the area, we first choose the small, horizontal element of area dA shown in Fig. 2. Approximating this area as a thin rectangle, the area is simply the product of the base times the height
<table cellSpacing=0 cellPadding=0 width=250 align=center border=0><tr vAlign=center align=middle><td>dA = w dy</TD>
<td align=right width=25>(2)</TD></TR></TABLE>
where w is the difference in the value of the x-coordinate at the right and left ends of the thin strip
<table cellSpacing=0 cellPadding=0 width=250 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(3)</TD></TR></TABLE>
(Note that the order in which we take the difference is important since the area and therefore w must be positive numbers.)
Then the area is
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(4)</TD></TR></TABLE>
which is about one-third the area of the bounding rectangle (4 × 8 = 32).
Finding the First Moment of Area - Mx
All of the horizontal strip of area in Fig. 2 is at the same y-coordinate. Therefore, the first moment of that little strip of area about the x-axis is just
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td>dMx = y dA</TD>
<td align=right width=25> </TD></TR></TABLE>
Adding the moments of all such horizontal strips together gives
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(5)</TD></TR></TABLE>
The y-coordinate of the centroid is then
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(6)</TD></TR></TABLE>
Finding the First Moment of Area - My
<table cellSpacing=0 cellPadding=5 align=right border=0><tr align=middle><td> </TD></TR>
<tr align=middle><td>Fig. 3</TD></TR></TABLE>
The horizontal strip of area in Fig. 2 is not all at the same x-coordinate, and it can not be used to calculate the first moment of that little strip about the y-axis
<table cellSpacing=0 cellPadding=0 width=250 align=center border=0><tr vAlign=center align=middle><td>dMy = x dA</TD>
<td align=right width=25> </TD></TR></TABLE>
Instead, we need to switch to a thin vertical strip of area as shown in Fig. 3. Again approximating the area as a thin rectangle, the area is simply the product of the height times the base
<table cellSpacing=0 cellPadding=0 width=250 align=center border=0><tr vAlign=center align=middle><td>dA = h dx</TD>
<td align=right width=25>(7)</TD></TR></TABLE>
where h is the difference in the value of the y-coordinate at the top and bottom ends of the thin strip
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>( </TD></TR></TABLE>
Adding the moments of all such vertical strips together gives
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(9)</TD></TR></TABLE>
and the x-coordinate of the centroid is then
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(10)</TD></TR></TABLE>
Sometimes the integrals obtained using one of the strips of area (say the vertical strip) are elementary, but the integrals obtained using the other strip of area (say the horizontal strip) are not elementary. In such cases, it would be nice to use the same strip of area (the vertical strip) for both calculations. This is the subject of the second example.
Centroid -
Method of Integration - 2
<table cellSpacing=0 cellPadding=5 width=600 align=center border=0><tr vAlign=top align=left><td bgColor=#4682b4> </TD>
<td>Example Problem
<table cellSpacing=0 cellPadding=5 align=right border=0><tr align=middle><td> </TD></TR>
<tr align=middle><td>Fig. 4</TD></TR></TABLE>
Use integration to locate the centroid of the shaded area shown in Fig. 4. </TD></TR></TABLE>
<table cellSpacing=0 cellPadding=5 width=600 align=center border=0><tr vAlign=top align=left><td bgColor=#4682b4> </TD>
<td>Solution
Finding the Area
<table cellSpacing=0 cellPadding=5 align=right border=0><tr align=middle><td> </TD></TR>
<tr align=middle><td>Fig. 5</TD></TR></TABLE>
To find the area, we first try the small, horizontal element of area shown in Fig. 5. Again approximating this area as a thin rectangle, the area is the product of the base times the height
<table cellSpacing=0 cellPadding=0 width=200 align=center border=0><tr vAlign=center align=middle><td>dA = w dy</TD>
<td align=right width=25> </TD></TR></TABLE>
where w is the difference in the value of the x-coordinate at the right and left ends of the thin strip
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(11)</TD></TR></TABLE>
But the integral of cos-1y is not one that I remember, and I would like to try something else if it would work.
<table cellSpacing=0 cellPadding=5 align=right border=0><tr align=middle><td> </TD></TR>
<tr align=middle><td>Fig. 6</TD></TR></TABLE>
Instead of the thin horizontal strip, we can use the thin vertical strip shown in Fig. 6. Again approximating the area of this strip as a thin rectangle the area is the product of the height times the base
<table cellSpacing=0 cellPadding=0 width=200 align=center border=0><tr vAlign=center align=middle><td>dA = h dx</TD>
<td align=right width=25> </TD></TR></TABLE>
where h is the difference in the value of the y-coordinate at the top and bottom ends of the thin strip
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(12)</TD></TR></TABLE>
Then the area is
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(13)</TD></TR></TABLE>
which is about two-thirds the area of the bounding rectangle (1 ×  /2).
Finding the First Moment of Area - My
All of the vertical strip of area in Fig. 6 is at the same x-coordinate. Therefore, the first moment of that little strip of area about the y-axis is just
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td>dMy = x dA</TD>
<td align=right width=25> </TD></TR></TABLE>Adding the moments of all such vertical strips together gives
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(14)</TD></TR></TABLE>The x-coordinate of the centroid is then
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(15)</TD></TR></TABLE>
Finding the First Moment of Area - Mx
We really ought to use the thin horizontal strip of area in Fig. 5 since all of the horizontal strip is at the same distance from the x-axis. However, the integral obtained using this approach is not elementary
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(16)</TD></TR></TABLE>and I would again like to find some way to use the thin vertical strip of Fig. 6 instead.
The first moment of an area about
<table cellSpacing=0 cellPadding=5 align=right border=0><tr align=middle><td> </TD></TR>
<tr align=middle><td>Fig. 7</TD></TR></TABLE>the x-axis is defined as
<table cellSpacing=0 cellPadding=0 width=200 align=center border=0><tr vAlign=center align=middle><td>dMx = y dA</TD>
<td align=right width=25> </TD></TR></TABLE>
Although all of the vertical strip of Fig. 6 is not at the same distance from the x-axis, we can use the centroid concept to concentrate all of the area of the thin vertical strip at its own centroid  = h /2 as shown in Fig. 7. Then the first moment of this concentrated element of area about the x-axis can be written
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(17)</TD></TR></TABLE>
Finally, adding the moments of all such vertical strips together gives
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(18)</TD></TR></TABLE>
and the y-coordinate of the centroid is then
<table cellSpacing=0 cellPadding=0 width=550 align=center border=0><tr vAlign=center align=middle><td> </TD>
<td align=right width=25>(19)</TD></TR></TABLE>
</TD></TR></TABLE>
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