6.3 Centroids (First moment of area)

6.3.1 First moment of the area about the y axis

Again consider curve y=f(x) from a to b, divided into strips of thickness δx. The area of the strip is given by (δA≈f((x))δx). The total area is given by the sum,

A≈∑baδA=∑baf(x)δx→∫baf(x)dx

If the strip is very thin then all of it is approximately at a distance x from y axis. If we now add up NOT δA but instead δA times x, i.e., δA “weighted” by x, we get the first moment of the area about the x axis,

Mx≈∑baxδA=∑baxf(x)δx→∫baxf((x))dx

This is usually called Mx, even though it is the first moment around the y axis.

Example 6.5: 

Find the first moment of area under y=1+x+x2 from x=0 to x=2 about the y axis.

Solution: 

Mx=====∫20x(1+x2+x3)dx∫20(x+x2+x3)dxx2∕2+x3∕3+x4∕4202+8∕3+426∕3

Example 6.6: 

Find the first moment of the area under y=e−x from x=0 to x=1 about the y axis.

Solution: 

Mx=∫10xe−xdx

Integrate by parts: u=x, du∕dx=1, du=dx, dv=e−xdx. Therefore dv∕dx=e−x, and thus v=−e−x,

Mx====xe−x10−∫10(−e−x)dx−e1−0+∫10e−xdx=−e1+(−e−x)10−e1+(−e1+1)1−e2=02642

6.3.2 First Moment of the area about the x axis

Figure 6.8: Subdividing the strips of width δx in ones of height δy.

Now consider the same strip of thickness δx. On this strip y goes from 0 to f(x). Divide strip into segments of length δy as shown in Fig. 6.8. The area of such a segment is δyδx. The total area of strip is δA≈∑f(x)y=0δyδx. In the limit that δy becomes infinitesimal we get

δA∫f(x)y=0dyδx==(y)0f(x)δxf(x)δx

as before. Now instead of summing segments we can weight each of them by the value of y to get

To find My we have to add the contributions of all strips

This is the formula for the first moment of the area about the x axis (This integral is same as that for the volume of revolution except for the factor 21 outside the integral rather than π).

Example 6.7: 

Find My for area under curve y=1+x+x2 from x=0 to x=2 (same area as in example xxxx(1))

Solution: 

f(x)f(x)2===1+x+x2(1+x+x2)21+2x+3x2+2x3+x4

Therefore

Missing or unrecognized delimiter for \bigg

6.3.3 Centroid of a plane area

Example 6.8: 

Find the centroid of the area under y=1+x+x2 from x=0 to x=2 using the previous two examples.

Solution: 

We know that Mx=26∕3 and My=71∕5, and we just need to determine A,

A====∫20(1+x+x2)dx(x+x2∕2+x3∕3)202+2+8∕320∕3

Therefore

xCyC==AMx=320326=2026=13AMy=320571=100213=213

6.3.4 Meaning of the centroid

If we have thin plate with constant thickness then the centroid is the position of centre of mass (C of M). The C of M is the point at which all mass can be regarded as acting. Let mass per unit area be ρ: This will be constant if the thickness is constant (and material is of uniform composition). The total mass m=Aρ where A is area. Turning effect about y axis of mass m at (xy) would be mx=Aρx. A strip of thickness δx, height f(x) has area f(x)δx. Mass would be ρf(x)δx. Total turning effect is ∑baxρf(x)δx→∫baxf(x)dx=ρMx, therefore Aρx=ρMx, therefore xC=1∕AMx.