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Chapter 6: Problem 6

Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the \(y\) -axis. $$y=\sqrt{x}, x=4, x=9, y=0$$

Short Answer

Expert verified
\(\frac{844\pi}{5}\) cubic units.

Step by step solution

01

Identify the region

Start by identifying the region bounded by the curves \(y=\sqrt{x}\), \(x=4\), \(x=9\), and \(y=0\). This region lies in the first quadrant where \(x\) ranges from 4 to 9 and \(y\) ranges from 0 up to \(y=\sqrt{x}\).
02

Understand the rotation

The region is revolved around the \(y\)-axis. Hence, we will use the method of cylindrical shells to find the volume of the solid generated by this rotation.
03

Set up the formula for volume using cylindrical shells

The volume \(V\) of the solid can be calculated using the cylindrical shell formula: \[ V = 2\pi \int_{a}^{b} x \, (f(x) - g(x)) \, dx \]Here, \(f(x)\) is \(\sqrt{x}\) (the upper boundary curve), \(g(x)\) is 0 (the lower boundary curve). The limits of integration are from \(x=4\) to \(x=9\).
04

Substitute into the formula

Substitute \(f(x)\) and \(g(x)\) into the volume integral: \[ V = 2\pi \int_{4}^{9} x \, (\sqrt{x} - 0) \, dx \]Simplify the expression to: \[ V = 2\pi \int_{4}^{9} x^{3/2} \, dx \]
05

Integrate the expression

Integrate the function \(x^{3/2}\): \[ \int x^{3/2} \, dx = \frac{2}{5} x^{5/2} + C \]Apply this to the integral: \[ V = 2\pi \left[ \frac{2}{5} x^{5/2} \right]_{4}^{9} \]
06

Evaluate the definite integral

Calculate the definite integral:1. Evaluate at the upper limit: \(\frac{2}{5} (9^{5/2})\) = \(\frac{2}{5} \times 243 = \frac{486}{5}\).2. Evaluate at the lower limit: \(\frac{2}{5} (4^{5/2})\) = \(\frac{2}{5} \times 32 = \frac{64}{5}\).3. Find the difference: \(\frac{486}{5} - \frac{64}{5} = \frac{422}{5}\).
07

Multiply by the constants

Multiply the result by \(2\pi\) to get the final volume:\[ V = 2\pi \times \frac{422}{5} = \frac{844\pi}{5} \]
08

Final answer

The volume of the solid generated is \(\frac{844\pi}{5}\).

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Key Concepts

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

Revolving Solids
Revolving solids is a fascinating topic in calculus that deals with creating three-dimensional shapes. This is done by rotating a two-dimensional area or curve around an axis. In this problem, the region defined by the curves is rotated about the \(y\)-axis which forms a 3D object called a solid of revolution.
Here's a simplified way to picture it:
  • Imagine a graph of the function \(y = \sqrt{x}\) between the lines \(x = 4\) and \(x = 9\).
  • This creates a flat, two-dimensional region under the curve.
  • When this region is spun around the \(y\)-axis, it forms a cylindrical shape, like a tunnel.
Understanding how to revolve solids, especially using different axes, opens the door to solving various engineering and physics problems. These skills are also foundational for disciplines that rely on spatial visualization and design.
Volume of Solids
Calculating the volume of solids is all about measuring the total space inside a 3D object. In the context of this exercise, once the shape is created by revolving the region, the next challenge is determining its volume. The method of cylindrical shells is the tool we'll use for this.
Cylindrical shells offer a unique way to observe and compute volume. Here's what makes them special:
  • They consider the solid as composed of thin, hollow cylinders called "shells." These shells wrap around the axis of rotation.
  • The height of each shell correlates with the value of the function being revolved, like \(y = \sqrt{x}\).
  • The thickness of each shell becomes an infinitesimally small piece on the \(x\)-axis, thus having height and radius accordingly.
Being familiar with calculating volumes using this method is crucial. It broadens the ability to handle different shapes and helps visualize how different parts contribute to the overall volume.
Definite Integrals
Definite integrals are essential tools for finding exact areas and volumes under curves, or within 3D shapes. They let us accumulate small pieces of area or volume over a defined interval.
In the context of finding volumes, definite integrals allow us to sum up the volumes of infinitesimally small shells, giving us the total volume of the solid.
  • The integral \(\int_{a}^{b} f(x) \, dx\) calculates the "total" accumulated from \(a\) to \(b\) of the function \(f(x)\).
  • In this problem, the limits \(x = 4\) to \(x = 9\) are used to define where the accumulation (or summing) takes place.
  • The function inside the integral represents the height multiplied by the circumference of each shell, using the formula \(2\pi\int_a^b x (f(x) - g(x)) \, dx\).
Definite integrals are fundamental because they precisely sum areas/volumes with exact bounds. This precision is vital in fields like engineering and physics, where accurate measurement and calculations are paramount.

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Most popular questions from this chapter

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