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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

Find the centroid of the region enclosed by the loop of the curve y2 = x3x4.

To determine

The centroid of the region enclosed by the curve y2=x3x4 .

Explanation

Given information:

The curve function is y2=x3x4 (1)

Calculation:

Rearrange Equation (1) as shown below:

y=±x3x4

Therefore, the loop of the curve have the functions of y=x3x4 and y=x3x4 .

Draw the loop of the curve for the function y2=x3x4 using the procedure as shown below:

  • Draw the graph for the function y=x3x4 by substituting different values for x.
  • Similarly in the same graph plot for the function y=x3x4 by substituting different values for x.

The region enclosed by the curves y=x3x4 and y=x3x4 is shown in Figure 1.

Refer Figure 1.

The loop of the curve is symmetric about x-axis. Therefore the y-coordinate of the centroid is y¯=0 .

The total area of the curve is twice the area of the top curve.

The interval of the loop of the curve is x=0 to x=1 .

The expression to find the area of the shaded region is shown below:

A=ab[f(x)]dx (2)

Here, the lower limit is a, the upper limit is b, and the top curve function is f(x) , and the bottom curve function is g(x) .

Rearrange Equation (2) for the curve being symmetric.

A=2ab[f(x)]dx (3)

Substitute 0 for a, 1 for b, and x3x4 for f(x) in Equation (3).

A=201(x3x4)dx=201(x321x)dx (4)

Let sinθ=x (5)

Find the limit of θ :

Substitute 0 for x in Equation (5).

sinθ=0θ=0

Substitute 1 for x in Equation (5).

sinθ=1sinθ=1θ=π2

Substitute sinθ for x and change the limits as 0toπ2 in Equation (4).

A=20π2(2sin4θcosθ1sin2θ)dθ=20π2(2sin4θcosθcos2θ)dθ=20π2(2sin4θcos2θ)dθ=40π2(14(1cos2θ)212(1+cos2θ))dθ

=120π2(1cos2θcos22θ+cos32θ)dθ=120π2(1cos2θ12(1+cos4θ)+cos2θ(1sin22θ))dθ=120π2(1cos2θ1212cos4θ+cos2θ(1sin22θ))dθ=120π2(1212cos4θcos2θsin22θ)dθ (6)

Integrate Equation (6) and apply the limits.

A=12[θ218sin4θ16sin32θ]0π2=12[(π418sin4(π2)16sin32(π2))(0)]=12(π40)=π8

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

x¯=1Aabx[f(x)]dx (7)

Rearrange Equation (7) for the curve being symmetric

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