   Chapter 8.3, Problem 42E

Chapter
Section
Textbook Problem

A rectangle with sides a and b is divided into two parts and by an arc of a parabola that has its vertex at one corner of and passes through the opposite corner. Find the centroids of both and . To determine

To find: The centroid of both part in the rectangle R1 and R2.

Explanation

Given:

The length of the rectangle is a.

The width of the rectangle is b.

Calculation:

Draw the rectangular as shown in Figure 1.

Refer to Figure 1.

The part in the rectangle (R1) is parabola‑1.

The part in the rectangle (R2) is parabola‑2.

Consider the area (A) of the rectangle (R) is,

A=ab

The equation of parabola is as shown,

y=kx2 (1)

Substitute a for x and b for y in Equation (1).

b=ka2k=ba2

Substitute ba2 for k in Equation (1).

y=ba2x2

Calculate the area (A1) of the parabola‑1:

A1=abydx (2)

Substitute 0 for a, a for b, and ba2x2 for y in Equation (2).

A1=0aba2x2dx=ba20ax2dx (3)

Integrate Equation (3).

A1=ba2[x2+12+1]0a=ba2[x33]0a=ba2(a330)=ab3

Calculate the (x¯1) coordinate of centroid of the parabola‑1:

x¯1=1A1abxydx (4)

Substitute 0 for a, a for b, ab3 for A1, and ba2x2 for y in Equation (4).

x¯1=1ab30ax(ba2x2)dx=3ab×ba20ax3dx=3a30ax3dx (5)

Integrate Equation (5).

x¯1=3a3[x3+13+1]0a=3a3[x44]0a=3a3(a440)=34a

Calculate the (y¯1) coordinate of centroid of the parabola‑1:

y¯1=1A1ab12[f(x)]2dx (6)

Substitute 0 for a, a for b, ab3 for A1, and ba2x2 for y in Equation (6).

y¯1=1ab30a12(ba2x2)2dx=3ab×12×b2a40ax4dx=3b2a50ax4dx (7)

Integrate Equation (7).

y¯1=3b2a5[x4+14+1]0a=3b2a5[x55]0a=3b2a5(a550)=310b

Hence, the centroid of part in the rectangle (R1) is (34a,310b)_.

Calculate the area (A2) of the parabola‑2:

A=A1+A2 (8)

Substitute ab for A, and ab3 for A1 in Equation (8)

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