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Shell method 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 indicated axis. \(y=\cos x^{2}, y=0,\) for \(0 \leq x \leq \sqrt{\pi / 2} ;\) about the \(y\) -axis

Short Answer

Expert verified
Answer: The volume of the solid generated is \(\pi\) cubic units.

Step by step solution

01

Define the volume of a cylindrical shell

In order to find the volume of the solid generated by revolving region R around the y-axis, we'll use the formula for the volume of a cylindrical shell: $$\Delta V = 2 \pi \cdot \text{radius} \cdot \text{height} \cdot \text{thickness}$$ For our vertical strips, the radius is \(x\), the height is \(\cos(x^2)\), and the thickness is \(\Delta x\).
02

Set up the integral for the volume

We'll integrate the volume of the shells across the given range of x-values, from \(0\) to \(\sqrt{\pi/2}\). The integral for the volume will be: $$V = \int_{0}^{\sqrt{\pi/2}} 2\pi \cdot x \cdot \cos(x^2) \, dx$$
03

Evaluate the integral

To evaluate the integral, we'll need to use a substitution method. Let's set \(u = x^2\) and \(du = 2x\,dx\). Now our integral becomes: $$V = \pi \int_{0}^{\pi/2} \cos(u) \, du$$ The integral of \(\cos(u)\) is \(\sin(u)\), so we have: $$V = \pi\left[\sin(u)\right]_{0}^{\pi/2}$$
04

Calculate the volume

Finally, we substitute the values back and calculate the volume: $$V = \pi\left[\sin \left(\frac{\pi}{2}\right) - \sin(0)\right] = \pi (1 - 0) = \pi$$ The volume of the solid generated when region R is revolved around the y-axis is \(\pi\) cubic units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solid of Revolution
The concept of a solid of revolution is a foundational idea in calculus that allows us to create a three-dimensional object by rotating a two-dimensional shape around an axis. Imagine taking a piece of paper with a shape drawn on it and spinning it around one of its edges—the shape sweeps out a volume in space.

For example, if we rotate a rectangle around one of its sides, we create a cylinder. In the exercise at hand, the region defined by the curve given by the equation \(y = \cos x^{2}\), the x-axis (\(y=0\)), and the vertical boundaries where \(0 \leq x \leq \sqrt{\pi / 2}\), is revolved around the y-axis to generate a more complex solid of revolution.

Understanding the properties of the shape's boundary is crucial as they determine the volume of the solid once it's revolved. It involves visualizing the shape as a collection of infinitesimally thin disks or shells which, when stacked together, form the complete volume. This leads to the application of the shell method to determine the total volume.
Volume of Cylindrical Shell
The volume of a cylindrical shell is found using a straightforward formula, intuitively derived from the geometry of cylinders. When a region is revolved around an axis, each tiny slice of the region becomes a 'shell'—think of rings of a tree or layers of an onion.

In calculus, we can slice the original region into vertical strips; when rotated, each strip forms a cylindrical shell. The volume \(\Delta V\) of a thin shell is given by the product of three factors:
  • Its circumference (\(2\pi\)) times the radius (\(x\)), which gives the shell's 'perimeter'
  • Its height (the value of the function, in this case, \(\cos(x^2)\))
  • Its thickness (\(\Delta x\), an infinitesimally small width)
Using these layers, we sum up all their volumes to determine the entire volume of the solid. The integral set up in the exercise uses this concept, integrating the volume of cylindrical shells from the innermost shell at \(x=0\) to the outermost shell at \(x=\sqrt{\pi/2}\).
Integration Techniques
To calculate the volume of the solid of revolution, we must evaluate the integral set up to represent the sum of all cylindrical shells' volumes. Integration techniques are various methods applied to solve integrals, which are the reverse operation of differentiation. Some common techniques include substitution, integration by parts, and partial fractions.

In this example, we use the substitution method to make the integral more easily solvable. The substitution \(u = x^{2}\) simplifies the integrand to \(\cos(u)\), and its antiderivative is \(\sin(u)\). We then evaluate this antiderivative at the boundaries to find the result.

Understanding the right technique to apply and performing the necessary algebraic manipulations is essential to calculate the volume correctly. It's not just about applying the formula mechanically; it's also about recognizing patterns, simplifying expressions, and accurately evaluating results. These skills are paramount to successfully using integration to solve real-world problems.

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Most popular questions from this chapter

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