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Chapter 6: Problem 11

Find the volume of the solid of revolution generated by revolving about the \(x\)-axis the region under each of the following curves. \(y=k x\) from \(x=0\) to \(x=h\) (generates a cone)

Short Answer

Expert verified
The volume is \( \frac{\pi k^2 h^3}{3} \).

Step by step solution

01

- Define the Solid of Revolution

The solid of revolution is obtained by revolving the region under the curve around the x-axis. Here, the curve is given by the equation: \(y = kx\). This needs to be revolved from \(x = 0\) to \(x = h\).
02

- Use the Volume Formula for Solids of Revolution

The volume \(V\) of a solid of revolution generated by revolving a function \(y = f(x)\) around the x-axis from \(x = a\) to \(x = b\) is given by the formula: \[V = \pi \int_a^b [f(x)]^2 dx\]
03

- Substitute the Given Function

Substitute the function \(y = kx\) into the volume formula. This gives: \[V = \pi \int_0^h (kx)^2 dx = \pi \int_0^h k^2 x^2 dx\]
04

- Integrate the Function

Integrate \(k^2 x^2\) with respect to \(x\): \[ \int_0^h k^2 x^2 dx = k^2 \int_0^h x^2 dx = k^2 \left[ \frac{x^3}{3} \right]_0^h = k^2 \left( \frac{h^3}{3} - \frac{0^3}{3} \right) = \frac{k^2 h^3}{3} \]
05

- Multiply by \(\pi\)

Multiply the result of the integral by \(\pi\) to find the volume: \[ V = \pi \frac{k^2 h^3}{3} = \frac{\pi k^2 h^3}{3} \]

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Key Concepts

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

integration
Integration is a fundamental concept in calculus. It helps in finding areas under curves, total accumulated quantities, and solving many real-world problems. In this context, integration is used to determine the volume of a solid of revolution. When we integrate a function, we are essentially summing up infinitesimally small slices of area to get a total volume. Each slice can be thought of as a thin disc. For the function given as part of this problem, integrating from 0 to h will sum up the areas of these discs to give us the total volume.
volume formula
The volume of a solid of revolution can be found using a specific formula. When a curve is revolved around the x-axis, the formula used is:


V =

蟺 鈭� [f(x)]虏 dx. Essentially, this formula converts the shape described by y = f(x) into a 3D object and calculates its volume by integrating the square of the function. The squaring accounts for the circular cross-sections of the shape. In our specific problem with the function y = kx revolved around the x-axis, we substitute kx into this formula and then perform the integration.
revolution around x-axis
When we talk about the revolution around the x-axis, we mean rotating a 2D region around the x-axis to create a 3D object. For example, in our problem, the line y = kx is rotated around the x-axis. This rotation creates a cone-like shape. Each point on the line traces a circle, and together, these circles form the solid. The radius of these circles varies with x, and that鈥檚 where the integration formula comes into play. We calculate the volume by summing up the areas of these circles from x = 0 to x = h.
cone volume
The volume of a cone can be derived using integration. For a cone generated by revolving the line y = kx around the x-axis from x = 0 to x = h, we use the integration method described earlier. After substituting y = kx into 蟺 鈭� [f(x)]虏 dx, integrating, and multiplying by 蟺, we ultimately get the formula for the volume of the cone: V = ( 蟺 k虏 h鲁 / 3 ). This is a specific instance of finding the volume of a solid of revolution and shows the power of calculus to derive formulas for complex shapes.

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