Problem 25 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\)-axis. \(y=\sqrt{\cos x}, \quad 0 \leq x \leq \pi / 2, \quad y=0, \quad x=0\)

Step-by-Step Solution

Answer

The volume of the solid is \(\pi\).

Step 1: Identify the Region

We need to determine the region that will be revolved about the x-axis. We have the curve \(y = \sqrt{\cos x}\) and the lines \(y = 0\), \(x = 0\), over the interval \(0 \leq x \leq \frac{\pi}{2}\). This forms a region above the x-axis bounded by these lines and the curve.

Step 2: Set Up the Integral for Volume

To find the volume of the solid formed by revolving the region around the x-axis, we use the disk method. The volume \(V\) is given by the integral:\[V = \pi \int_{0}^{\frac{\pi}{2}} (\sqrt{\cos x})^2 \, dx\]This integral represents the sum of an infinite number of disks with radius \(\sqrt{\cos x}\).

Step 3: Simplify the Integrand

Simplify the integrand \((\sqrt{\cos x})^2\) as follows:\[(\sqrt{\cos x})^2 = \cos x\]Thus, the integral becomes:\[V = \pi \int_{0}^{\frac{\pi}{2}} \cos x \, dx\]

Step 4: Calculate the Integral

Evaluate the integral \(\pi \int_{0}^{\frac{\pi}{2}} \cos x \, dx\). The antiderivative of \(\cos x\) is \(\sin x\):\[\pi \left[\sin x \right]_{0}^{\frac{\pi}{2}} = \pi (\sin(\frac{\pi}{2}) - \sin(0))\]Calculating the values gives \(\sin(\frac{\pi}{2}) = 1\) and \(\sin(0) = 0\).

Step 5: Final Calculation

Substitute the calculated values into the expression from the antiderivative:\[\pi (1 - 0) = \pi\]Thus, the volume of the solid is \(\pi\).

Key Concepts

Disk Method

The Disk Method is an essential technique used in calculus to find the volumes of solids of revolution. When a region in the plane is revolved around an axis, it can produce a three-dimensional solid. To compute the volume of such a solid, the Disk Method divides the shape into thin disks. Each disk has a small thickness, usually represented by a differential element, such as \(dx\).

• Each disk’s radius is determined by the function that defines the curve.
• The disk's volume is approximately the area of its circular face multiplied by its thickness, \( V \approx \pi r^2 dx \).
• The integral combines all these tiny volumes together from one endpoint of the region to the other.

By imagining these disks stacked along the axis of revolution, the integral sums their volumes to obtain the total volume of the solid. This method is particularly effective when rotating around the horizontal or vertical axis.

Integral Calculus

Integral Calculus is a branch of calculus that focuses on accumulation and the calculation of areas under curves. It is the opposite of differential calculus, which concerns itself with rates of change.

• The Fundamental Theorem of Calculus connects derivatives with integrals, showing that integration can undo differentiation.
• An integral, often represented by the symbol \( \int \), is used to calculate a plethora of quantities such as area, volume, and other accumulative measures.
• Definite integrals compute the total accumulation over a specific interval, as shown in the original exercise, where it determines the volume under a curve, generating a solid when revolved around an axis.

Integral Calculus provides the tools to evaluate these integrals by considering the function's antiderivative.

Antiderivative

An Antiderivative of a function is another function whose derivative gives back the original function. Finding an antiderivative is the process of integration in calculus and allows us to solve a variety of integral problems.

• If \(F(x)\) is an antiderivative of \(f(x)\), then \(F'(x) = f(x)\).
• In the context of the original exercise, the antiderivative of \(\cos x\) is \(\sin x\).
• This relationship enables you to evaluate definite integrals by calculating the difference between the antiderivative's values at the given limits.

The antiderivative is a crucial step in solving volumes using integral calculus as it transitions us from an indefinite to a definite integral.

Cosine Function

The Cosine Function, \(\cos x\), is one of the fundamental trigonometric functions. It expresses the relation of the angle in a right triangle with the ratio of the length of the adjacent side to the hypotenuse.

• It is periodic with a period of \(2\pi\), meaning \(\cos(x + 2\pi) = \cos x\).
• The function oscillates between -1 and 1, reaching its maximum value at certain points, like \(\cos(0) = 1\).
• In calculus, the derivative of \(\cos x\) is \(-\sin x\), and its antiderivative is \(\sin x\).

In the provided exercise, \(\cos x\) forms the basis of calculating the volume. It appears in the integrand within the function under the square root, showcasing its role in defining the boundary of the region.