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 \(x\) -axis. $$x=y^{2}, x=4, \text { and } y=0$$

Short answer

Question: Determine the volume of the solid generated by revolving the region R bounded by the curves \(y = \cos(x)\), \(y=0\), and \(x=0\) about the x-axis using the disk method. Solution: The volume of the solid generated by revolving region R around the x-axis is \(\frac{\pi^2}{4}\).

Step-by-step solution

Identify the bounds, the functions, and the axis of revolution

The given problem involves the region R is bounded by the following curves: 1. \(y = \cos(x) \) 2. \(y = 0\) 3. \(x = 0\) The axis of revolution is the x-axis. Now let's determine the interval bounds over which the region R exists. The function \(y = \cos(x)\) has a period of \(2\pi\), and since the lower bound is \(x = 0\), the upper bound will be the first intersection point of the curve with the x-axis, when \(\cos(x) = 0\). Solving for x, we get \(x=\frac{\pi}{2}\). Thus the interval is [0, \(\frac{\pi}{2}\)].

Calculate the volume of each slice using the cross-sectional area of each disk

We will use the disk method to calculate the volume of each slice with respect to the x-axis. The cross-sectional area A of each disk can be calculated using the formula: $$A(x) = \pi \cdot [f(x)]^2$$ Here, the function f(x) is the distance from the curve y = cos(x) to the x-axis. Since the x-axis is the axis of revolution, the distance from the curve to the x-axis is equal to y itself which is the function \(f(x) = \cos(x)\). Now, we will plug in the function f(x) into the formula for the cross-sectional area: $$A(x) = \pi(\cos(x))^2$$

Integrate the volumes over the given interval to find the total volume

Now we have to integrate A(x) over the interval [0, \(\frac{\pi}{2}\)] to find the volume: $$V = \int_{0}^{\frac{\pi}{2}} A(x) dx = \int_{0}^{\frac{\pi}{2}} \pi(\cos(x))^2 dx$$ Now we can use the identity that was provided in the problem statement: \(\cos^2(x) = \frac{1}{2}(1+\cos(2x))\). Substitute this into the integral: $$V = \pi \int_{0}^{\frac{\pi}{2}} \frac{1}{2} (1 + \cos(2x)) dx$$ Now integrate: $$V = \pi \left[ \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) dx \right]$$ $$V = \pi \left[ \frac{1}{2} \left( \left[ x \right]_{0}^{\frac{\pi}{2}} \right) + \frac{1}{2} \left( \left[ \frac{1}{2}\sin(2x) \right]_{0}^{\frac{\pi}{2}} \right) \right] $$ Evaluate the integral: $$V = \pi \left( \frac{1}{2} \cdot \frac{\pi}{2} + \frac{1}{2} \cdot (0) \right) = \frac{\pi^2}{4}$$ So the volume of the solid generated by revolving R about the x-axis is \(\frac{\pi^2}{4}\).

Key concepts

Volume of Solids of Revolution

When we talk about the volume of solids of revolution, we're exploring a fascinating way to create 3D shapes from 2D regions. Imagine taking a flat region on a graph and spinning it around a line, like the x-axis. This creates a solid object, sort of like rotating a semicircle to form a sphere.

• To find the volume of such a solid, we use methods like the disk and washer methods.
• The disk method involves slicing the solid perpendicular to the axis of rotation, giving us tiny disk-like shapes.
• We calculate the volume of each disk and then sum them up using integration across the specified interval.

For this specific exercise, because the region is revolved around the x-axis, each slice or disk is of a height (or thickness) equivalent to a tiny change in x. The radius of the disk is determined by the distance from the curve, here given by the cosine function, to the axis of rotation.

Definite Integral

The definite integral is a powerful tool in calculus used to calculate the accumulation of quantities, like areas under curves or volumes of solids. When dealing with volumes of revolution, the definite integral allows us to aggregate the volume of many tiny slices to find the total volume.

• The integral is expressed with limits that define the range of integration, ensuring we only account for the relevant section of the curve.
• In our exercise, the bounds are from 0 to \(\frac{\pi}{2}\), representing the interval where the cosine function lies above the x-axis.
• By setting up the integral of the function that represents the area of each disk, \(\pi [\cos(x)]^2\), we can compute the total volume.

The definite integral sums the areas of all the disks from the start to the end of the interval, giving us the volume of the solid formed.

Trigonometric Functions

Trigonometric functions are the functions of angles and have vast applications in various fields, including calculus. Functions like the cosine, sine, and tangent help describe periodic phenomena and can also describe the boundary of regions being revolved to form solids.

• The cosine function, \(y = \cos(x)\), gives a wave-like shape that repeats every \(2\pi\) radians.
• In the context of this problem, \(\cos(x)\) is the curve being revolved around the x-axis to construct our 3D shape.
• To simplify integration, identities like \(\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\) are used.

These identities transform the integrand into a more manageable form, making the final calculation of the integral straightforward and demonstrating the versatility and power of trigonometry in calculus applications.