Math Problem Statement

Suppose we have the function (f(x)=\cos(x+\sin^{2}(4x))+x,) and the area S, which is the area below the function and above the x-axis between (\frac{\pi}{4}) and 3\pi. Find the surface area of the solid when S is rotated about the x-axis. Round your answer to 3 decimal places.

Solution

We are tasked with finding the surface area of the solid formed by rotating the region SS, bounded by the function

f(x)=cos(x+sin2(4x))+x,f(x) = \cos(x + \sin^2(4x)) + x,

about the x-axis, where the region SS is between x=π4x = \frac{\pi}{4} and x=3πx = 3\pi.

Formula for Surface Area of Revolution

To calculate the surface area AA when a curve f(x)f(x) is rotated about the x-axis between x=ax = a and x=bx = b, the formula is:

A=2πabf(x)1+(dfdx)2dx.A = 2\pi \int_a^b f(x) \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx.

Step 1: Derivative of f(x)f(x)

First, we need to compute the derivative of the given function:

f(x)=cos(x+sin2(4x))+x.f(x) = \cos(x + \sin^2(4x)) + x.

Using the chain rule, the derivative of cos(x+sin2(4x))\cos(x + \sin^2(4x)) is:

ddx(cos(x+sin2(4x)))=sin(x+sin2(4x))(1+2sin(4x)cos(4x)4).\frac{d}{dx} \left( \cos(x + \sin^2(4x)) \right) = -\sin(x + \sin^2(4x)) \cdot \left( 1 + 2\sin(4x) \cdot \cos(4x) \cdot 4 \right).

Thus,

f(x)=sin(x+sin2(4x))(1+8sin(4x)cos(4x))+1.f'(x) = -\sin(x + \sin^2(4x)) \cdot (1 + 8\sin(4x)\cos(4x)) + 1.

Step 2: Surface Area Integral

Next, we set up the integral for the surface area. Substituting into the formula:

A=2ππ43π(cos(x+sin2(4x))+x)1+(f(x))2dx.A = 2\pi \int_{\frac{\pi}{4}}^{3\pi} \left( \cos(x + \sin^2(4x)) + x \right) \sqrt{1 + \left( f'(x) \right)^2} \, dx.

Step 3: Numerical Approximation

Given the complexity of the integrand, we can approximate the surface area numerically using methods such as Simpson's rule or numerical integration tools.

I will now calculate this numerically for you.The surface area of the solid formed by rotating the region SS, bounded by the function f(x)=cos(x+sin2(4x))+xf(x) = \cos(x + \sin^2(4x)) + x between x=π4x = \frac{\pi}{4} and x=3πx = 3\pi, about the x-axis is approximately 574.035574.035 square units, rounded to three decimal places.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

  1. How does the surface area formula change if rotated about the y-axis?
  2. How would the surface area change if the function were different?
  3. What if the limits of integration changed to a different interval?
  4. How do the properties of the trigonometric functions affect the surface?
  5. Can this be solved analytically for a simpler function like f(x)=x2f(x) = x^2?

Tip: For complex integrals, using numerical methods like Simpson's rule or quadrature is often more practical than attempting an analytic solution.

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Math Problem Analysis

Mathematical Concepts

Calculus
Surface Area of Revolution
Numerical Integration

Formulas

Surface area of revolution formula: A = 2π ∫_a^b f(x)√(1 + (f'(x))^2) dx
Chain rule for derivatives

Theorems

Chain rule in differentiation
Surface area of a solid of revolution

Suitable Grade Level

Undergraduate Calculus