Problem 10 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. \(y=1-x^{2}, x=0,\) and \(y=0,\) in the first quadrant

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{1}{2}\pi\).

Step 1: Sketch the region bounded by the curves

Begin by sketching the region in the first quadrant that is bounded by the curves \(y=1-x^2\), \(x=0\), and \(y=0\). You should see a parabolic region, with its vertex at the point \((0,1)\), pointing downwards and opening along the positive \(x\)-axis.

Step 2: Set up the volume integral using the shell method

To find the volume using the shell method, we will consider an infinitesimally thin shell of thickness \(dx\) at a distance \(x\) from the \(y\)-axis. The height of the shell will be given by the function \(y=1-x^2\). The circumference of the shell is \(2\pi x\) and its volume is given by \(2\pi x(1-x^2)dx\). To find the total volume of the solid, we will sum up the volumes of these shells, integrating along the \(x\)-axis from \(0\) to \(x=1\) (the upper limit comes from solving the equation \(1-x^2=0\) for \(x\)). We will calculate the volume using the integral: $$V = \int_{0}^{1} 2\pi x(1-x^2) dx$$

Step 3: Evaluate the integral to find the volume of the solid

Now, we will evaluate the integral to find the total volume of the solid generated: $$V = \int_{0}^{1} 2\pi x(1-x^2) dx = 2\pi \int_{0}^{1} (x - x^3) dx = 2\pi \left[\frac{x^2}{2} - \frac{x^4}{4} \right]_0^1$$ Plug in the limits of integration: $$V = 2\pi \left[\frac{(1)^2}{2} - \frac{(1)^4}{4} - \frac{(0)^2}{2} + \frac{(0)^4}{4} \right]= 2\pi \left[\frac{1}{2} - \frac{1}{4}\right] = 2\pi \left[\frac{1}{4}\right]$$ Finally, we get the volume of the solid: $$V = \boxed{\frac{1}{2}\pi}$$

Key Concepts

Volume of Revolution

When you revolve a 2-dimensional shape around an axis, it creates a 3-dimensional object. This process is called finding the volume of revolution. In this context, using the shell method allows us to determine the volume of a solid of revolution in a more intuitive manner.

The shell method involves using cylindrical shells to approximate the volume. Imagine slicing the region into many thin vertical slices parallel to the axis of revolution. Each slice will contribute a small, thin cylindrical shell to the total volume. By summing all these tiny shells, we can compute the total volume of the solid.

This method is particularly useful when dealing with rotating a region around an axis that is not parallel to the shapes' natural layout. A typical setup involves integrating the surface area of these shells over a specific interval.

Integral Calculus

Integral calculus is a branch of mathematics that deals with accumulations, such as areas under curves and totals of infinite series. It is particularly useful in solving problems related to volumes of solid figures. For finding volumes of revolution, integral calculus allows us to sum up an infinite number of infinitesimally small quantities to get precise volume calculations.

In this exercise, we use an integral to sum the volumes of all the cylindrical shells formed by revolving the region around an axis. The formula for shell volume involves integrating along the defined limits of the region. Solving the integral helps in expressing the volume as a concrete numerical value.

Using calculus and understanding how to set up integrals are essential skills for calculating areas and volumes in advanced mathematics. Techniques such as substitution can also be involved, which simplify complex integral evaluations.

Parabolas

Parabolas are u-shaped curves defined by quadratic equations like the one in this problem, which is given by \(y = 1 - x^2\). In this specific region, the parabola opens downward with its vertex at the point (0, 1). The parabola intersects the x-axis at points derived by setting \(1-x^2 = 0\), providing limits for integration.

Understanding the properties of parabolas is crucial when determining the boundaries of the region to be revolved. These curves are symmetric about their axis, and their steepness or width depends on the coefficient in front of \(x^2\).

Here, the parabola helps define the upper limit of the region, and working with it accurately ensures the right boundaries are chosen for computing the volume. Mastery of parabola characteristics assists in solving range and area problems, which are common in calculus applications.