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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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Section
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

Find the centroid of the region shown.

14. images

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+14πr2 (1)

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

A=(12×8×8)+(14π(8)2)=32+16π=16(2+π)

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=8m=1

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

y=x+8

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

y=r2x2 (3)

Here, the radius of quarter circle is r.

Substitute 8 for r in Equation (3).

y=82x2

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

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

x¯=1Aabx[f(x)g(x)]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+π) for A, x+8 for f(x) , and 82x2 for g(x) in Equation (4).

x¯=116(2+π)08x[(x+8)+82x2]dx=116(2+π)08[x2+8x+x64x2]dx=116(2+π)08[x2+8x+x(64x2)12]dx=116(2+π)[08x2dx+088xdx+08x(64x2)12dx] (5)

Find the value of the integral 08x2dx :

08x2dx=[x33]08=(833)(0)=5123

Find the value of the integral 088xdx :

088xdx=[8x22]08=(4(8)2)(0)=256

Find the value of the integral 08x(64x2)12dx :

Let u=64x2 (6)

Differentiate Equation (6) with respect to x.

du=2xdxdx=du2x

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

08x(64x2)12dx=08x(u)12(du2x)=1208(u)12du=12[u3232]08=13[u32]08 (7)

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

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