Chapter 5: Applications of Integration
Section 5.7: Centroids
Example 5.7.5
Determine the centroid of the trapezoid 25 × 15 × 15 × 15 (feet), longest edge uppermost and horizontal.
Solution
Mathematical Solution
By the Pythagorean theorem and Figure 5.7.5(a), the height h of the trapezoid is 152−52=102.
With respect to Figure 5.7.5(a), let point P be the origin of a Cartesian coordinate system. Then, the coordinates of points P,Q,R,S, are respectively,
0,0,−5,h,20,h,15,0
The left and right edges of the trapezoid are given respectively by the equations
yL=−22 x and yR=22x−15
The area of the trapezoid is
A=15+252 102=2002
Figure 5.7.5(a) 25 × 15 × 15 × 15 trapezoid
If x&conjugate0;,y&conjugate0; is the centroid, then x&conjugate0;=7.5 (by symmetry) and
y&conjugate0; = 12 A∫−50h2−yL2 ⅆx+∫015h2 ⅆx+∫1520h2−yR2 ⅆx=652/12 ≐ 7.66
Maple Solution
Initialize
Tools≻Load Package: Student Precalculus
Loading Student:-Precalculus
Define h,P,Q,R,S
Context Panel: Assign Name
h=102→assign
P=0,0→assign
Q=−5,h→assign
R=20,h→assign
S=15,0→assign
Area of the trapezoid
A=15+252h→assign
Equations of the left and right edges
Apply the Line command from the Student Precalculus package.
LineP,Q1;LineR,S1
y=−2⁢2⁢x
yL=−2⁢2⁢x→assign
yR=22x−15→assign
Apply the formula y&conjugate0;=1A∫abf2x−g2x/2 ⅆx from Table 5.7.1
12 A∫−50h2−yL2 ⅆx+∫015h2 ⅆx+∫1520h2−yR2 ⅆx = 6512⁢2→at 10 digits7.660323461
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