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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then use your calculator to find (approximately) the volume of the solid obtained by rotating about the x-axis the region bounded by these curves.

y = ln(x6 + 2), y = 3 x 3

To determine

To calculate: The approximate x-coordinates of the point of intersection of the given curves and the volume of the solid obtained by rotating about the x-axis.

Explanation

Given information:

The Equation of the curves y=ln(x6+2) and y=3x3.

Calculation:

Show the Equation of the curves as follows:

y=ln(x6+2) (1)

y=3x3 (2)

Procedure to sketch the region bounded by the two curves is explained below:

  • Draw the graph for the function y=ln(x6+2) by substituting different values for x.
  • Similarly plot for the function y=3x3 by substituting different values for x.

Sketch the region enclosed by the curves as shown in Figure 1.

Find the intersection points of the curves as shown below.

Refer to Figure 1.

Consider that the intersection points are x=4.09, x=1.47 and x=1.1.

Hence, the approximate x-coordinates of the point of intersection of the given curves are x=4.09_, x=1.47_ and x=1.1_.

Refer to Figure 1.

Consider that the outer radius of the region r1=ln(x6+2) and the inner radius of the region r2=3x3 within the limit 4.09 to 1.47.

Consider that the outer radius of the region r1=3x3 and the inner radius of the region r2==ln(x6+2) within the limit 1.47 to 1.1.

Find the Area enclosed by the curves as shown below.

A2(x)=πr12πr22=π(ln(x6+2))2π(3x3)2=π((ln(x6+2))23+x3)

A2(x)=πr12πr22=π(3x)2π(ln(x6+2))2=π(3x3(ln(x6+2))2)

The expression to find the volume as shown below.

V=abA(x)dx

Find the volume of the solid obtained by rotating about the x-axis as shown below.

V=a1b1A1(x)dx+a2b2A2(x)dx (3)

Substitute 4.09 for a1, 1.47 for b1, 1.47 for a2, 1.1 for b2,π((ln(x6+2))23+x3) for A1(x), and π(3x3(ln(x6+2))2) for A2(x) in Equation (3)

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