Chapter 8, Problem 14RE

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

# Find the centroid of the region shown.14.

To determine

To find: The centroid of the shaded region.

Explanation

Given:

The width and height of the triangle are 8, 8.

Radius of the Quarter circle is 8.

Calculation:

The area of the shaded region is obtained by adding the area of triangle and the area of the quarter circle.

Calculate the area of the shaded region (A) using the graph:

A=12bh+14Ï€r2 (1)

Substitute 8 for b, 8 for h, and 8 for r in Equation (1).

A=(12Ã—8Ã—8)+(14Ï€(8)2)=32+16Ï€=16(2+Ï€)

The general expression for an equation of the line is shown below:

y=mx+C (2)

Here, the slope is m and the constant is C.

Substitute 8 for y and 0 for x in Equation (2).

8=0+CC=8

Substitute 0 for y, 8 for x, and 8 for C in Equation (2).

0=m(8)+88m=âˆ’8m=âˆ’1

Substitute âˆ’1 for m and 8 for C in Equation (2).

y=âˆ’x+8

The general expression for an equation of the quarter circle is shown below:

y=r2âˆ’x2 (3)

Here, the radius of quarter circle is r.

Substitute 8 for r in Equation (3).

y=82âˆ’x2

The circle lies in the third quadrant, thus the equation of the quarter circle is y=âˆ’82âˆ’x2 .

Calculate the x-coordinate of the centroid (xÂ¯) using the relation:

xÂ¯=1Aâˆ«abx[f(x)âˆ’g(x)]â€‰dx (4)

Here, the lower limit is a, the upper limit is b and the functions are f(x) , g(x) .

Substitute 0 for a, 8 for b, 16(2+Ï€) for A, âˆ’x+8 for f(x) , and 82âˆ’x2 for g(x) in Equation (4).

xÂ¯=116(2+Ï€)âˆ«08x[(âˆ’x+8)+82âˆ’x2]â€‰dx=116(2+Ï€)âˆ«08[âˆ’x2+8x+x64âˆ’x2]â€‰dx=116(2+Ï€)âˆ«08[âˆ’x2+8x+x(64âˆ’x2)12]â€‰dx=116(2+Ï€)[âˆ«08âˆ’x2â€‰dx+âˆ«088xâ€‰dx+âˆ«08x(64âˆ’x2)12â€‰dx] (5)

Find the value of the integral âˆ«08âˆ’x2â€‰dx :

âˆ«08âˆ’x2â€‰dx=[âˆ’x33]08=(âˆ’833)âˆ’(0)=âˆ’5123

Find the value of the integral âˆ«088xâ€‰dx :

âˆ«088xâ€‰dx=[8x22]08=(4(8)2)âˆ’(0)=256

Find the value of the integral âˆ«08x(64âˆ’x2)12â€‰dx :

Let u=64âˆ’x2 (6)

Differentiate Equation (6) with respect to x.

du=âˆ’2xdxdx=âˆ’du2x

Substitute u for 64âˆ’x2 and âˆ’du2x for dx in the integral.

âˆ«08x(64âˆ’x2)12â€‰dx=âˆ«08x(u)12â€‰(âˆ’du2x)=âˆ’12âˆ«08(u)12â€‰du=âˆ’12[u3232]08=âˆ’13[u32]08 (7)

Substitute 64âˆ’x2 for u and apply the limits of x in Equation (7)

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