Question

FInd the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(y=\sqrt{x}, x=5, y=0 ;\) about the line \(x=5\)

Short answer

The volume is approximately 16.43\( \pi \).

Step-by-step solution

Sketch the Region R

Sketch the region bounded by the curves. The curve is given by \( y = \sqrt{x} \), from \( x = 0 \) to \( x = 5 \), and \( y = 0 \) which is the x-axis. The region is the area under \( y = \sqrt{x} \) and above the x-axis from \( x = 0 \) to \( x = 5 \).

Draw and Label a Typical Rectangular Slice

Choose a vertical slice at any arbitrary x between 0 and 5. The height of this slice is given by \( y = \sqrt{x} \), and the slice's width is \( dx \). When revolved about the line \( x = 5 \), this slice forms a cylindrical shell.

Formula for Approximate Volume

The formula for the volume of a cylindrical shell is \( 2\pi \times \text{radius} \times \text{height} \times \text{thickness} \). Here, the radius is \( 5 - x \), the height is \( \sqrt{x} \), and the thickness is \( dx \). So, the approximate volume of the shell is \( 2\pi(5 - x)\sqrt{x}\,dx \).

Set Up the Integral

To find the total volume, integrate the approximate volume expression from \( x = 0 \) to \( x = 5 \). The integral becomes: \[ V = \int_{0}^{5} 2\pi(5 - x)\sqrt{x}\, dx \]

Evaluate the Integral

Expand and evaluate the integral: \( V = 2\pi \int_{0}^{5} (5\sqrt{x} - x\sqrt{x})\, dx \)Calculate each part separately:- \( \int 5\sqrt{x}\, dx = \frac{10}{3}x^{3/2} \)- \( \int x^{3/2}\, dx = \frac{2}{5}x^{5/2} \)Substitute back:\[ V = 2\pi \left[ \frac{10}{3}x^{3/2} - \frac{2}{5}x^{5/2} \right]_{0}^{5} \]Evaluate:\[ V = 2\pi \left( \left[ \frac{10}{3}(5)^{3/2} - \frac{2}{5}(5)^{5/2} \right] - \left[ \frac{10}{3}(0)^{3/2} - \frac{2}{5}(0)^{5/2} \right] \right) \]Calculate values and simplify to find \( V \).

Calculate Numerical Answer

Substitute the calculated integrals values into\( V = 2 \pi \left[ \frac{10}{3}(5)^{3/2} - \frac{2}{5}(5)^{5/2} \right] \)Solve:- Calculate \( 5^{3/2} = \sqrt{125} \)- Calculate \( 5^{5/2} = \sqrt{3125} \)Evaluate each expression numerically and complete the simplification to find the final answer for \( V \), which is approximately 16.43\( \pi \).

Key concepts

Cylindrical Shell Method

The cylindrical shell method is a creative and efficient way to find the volume of a solid of revolution.
It involves revolving a region around an axis to create a 3D object. In this method, you imagine slicing the region into thin vertical slices.
Upon revolving these slices, they form cylindrical shells. Each slice adds to the total volume of the solid. Think of each slice as a thin, hollow cylinder.
The volume of this cylinder is calculated by multiplying the circumference of its base, its height, and its thickness. In mathematical terms, for a slice at position \(x\),

• The radius of the cylinder is \(5 - x\) (distance from the line of revolution \(x = 5\)).
• The height of the cylinder is \(\sqrt{x}\) (height of the region).
• The thickness is \(dx\) (an infinitesimal increment along the \(x\)-axis).

To sum up all these individual shell volumes, you set up an integral that covers the entire range of the region.

Integral Calculus

Integral Calculus is like the mathematical glue that holds the cylindrical shell method together.
The core idea is to use integration to sum up an infinite number of infinitesimally small parts to find quantities like area and volume. In this technique, calculating the volume of the solid uses the integral of the formula for the cylindrical shell we derived earlier.
You set up the integral to cover the entire region over which the solid is formed:\[V = \int_{0}^{5} 2\pi (5 - x)\sqrt{x}\, dx\]This integral expression represents the total volume of the solid from revolving the region bounded by the curves from \(x = 0\) to \(x = 5\).
The process of solving this integral involves determining primitives, substitution, and boundary conditions, bringing solutions to these seemingly complex problems.

Calculus Problem Solving

Calculus problem solving is both an art and a science. It involves using mathematical principles and techniques to explore and solve problems.
In the case of calculating volumes of revolution, problem-solving might seem daunting at first. Here’s a breakdown to ease the process:

• Understand the Problem: Grasp what is being asked, such as finding the volume of a solid obtained from revolving an area.
• Visualize: Drawing the region or solid helps comprehend what you are dealing with.
• Set Up Your Equations: Use the right formula, in this case, the cylindrical shell formula derived methodologically.
• Integrate: Solve the integral with respect to your limits; not just mechanically but also understanding each step.
• Check: Verify your calculations with boundary conditions to ensure accuracy.

By practicing these steps, you gradually become adept at tackling even more complex calculus problems.

Definite Integral

A definite integral is an essential tool in calculus, giving us a way to calculate the exact value of an expression over a specific interval.
It not only helps in calculating areas under curves but also the volumes of solids, as seen in the volume of revolution problems. The definite integral has upper and lower limits, indicating where the integration starts and ends. In this exercise, the integral:\[V = \int_{0}^{5} 2\pi (5 - x)\sqrt{x}\, dx\]is 'definite' because it is evaluated from \(x = 0\) to \(x = 5\).
Solving this requires calculating antiderivatives, applying the limits, and sometimes simplifying expressions for the final numeric solution.Each step in definite integration presents opportunities to understand calculus principles deeply. Many students often find the transition from setting up an integral to calculating the final number as illuminating, showcasing calculus's power and potential in real-life applications.