Question

88. The region in Exercise 87 is revolved about the x-axis to generate a solid.b. Show that the inner and outer surfaces of the solid have infinite area.

Accepted Answer

Recall that the problem involves revolving a region about the x-axis to generate a solid. The surface area of a solid of revolution about the x-axis is given by the formula:

$$ S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ Identify the functions that define the inner and outer surfaces of the solid from Exercise 87. These functions describe the boundaries of the region being revolved. Let’s denote them as $$y = f(x)$$ for the outer surface and $$y = g(x)$$ for the inner surface, where $$f(x) \geq g(x) \geq 0 on $$the interval $$[a,b]. $$Compute the derivatives $$\frac{df}{dx}$$ and $$\frac{dg}{dx}$$ for the outer and inner functions respectively. These derivatives are necessary to evaluate the integrand $$\sqrt{1 + \left(\frac{dy}{dx}\$$right)^2$$} in $$the surface area formula. Set up the surface area integrals for both the outer and inner surfaces using the formula:

$$ S_{outer} = \int_a^b 2\pi f(x) \sqrt{1 + \left(\frac{df}{dx}\right)^2} \, dx $$

$$ S_{inner} = \int_a^b 2\pi g(x) \sqrt{1 + \left(\frac{dg}{dx}\right)^2} \, dx $$ Analyze the behavior of the integrands near any points where the functions or their derivatives become unbounded or approach zero in a way that causes the integral to diverge. Show that these integrals do not converge to a finite value, thus proving that the inner and outer surfaces have infinite area.

88. The region in Exercise 87 is revolved about the x-axis to generate a solid.
b. Show that the inner and outer surfaces of the solid have infinite area.

Verified video answer for a similar problem:

Key Concepts

Surface Area of a Solid of Revolution

Improper Integrals and Infinite Areas

Behavior of Functions Near Singularities or Infinity