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Question Let \( R \) represent the region bounded above by the graph of \( f(x)=2 x+7 \), below by the graph of \( g(x)=\frac{6}{x} \), and on eac side by \( x=1 \) and \( x=4 \). What is the volume of the solid of revolution obtained by rotating \( R \) about the \( x \)-axis? Enter an exact value in terms of \( \pi \).

Ask by Bird Hammond. in the United States
Jan 19,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The volume of the solid of revolution is \( 414\pi \) cubic units.

Solution

To find the volume of the solid of revolution obtained by rotating the region \( R \) about the \( x \)-axis, we can use the method of disks. The volume of the solid of revolution is given by the formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 - [g(x)]^2 \, dx \] where \( f(x) \) is the function that bounds the region above, \( g(x) \) is the function that bounds the region below, and \( a \) and \( b \) are the limits of integration. In this case, \( f(x) = 2x + 7 \) and \( g(x) = \frac{6}{x} \). The limits of integration are \( a = 1 \) and \( b = 4 \). Substitute the functions and limits into the formula: \[ V = \pi \int_{1}^{4} (2x + 7)^2 - \left(\frac{6}{x}\right)^2 \, dx \] Now, we need to calculate the integral to find the volume of the solid of revolution. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\pi \times \int_{1}^{4} \left(\left(2x+7\right)^{2}-\left(\frac{6}{x}\right)^{2}\right) dx\) - step1: Subtract the terms: \(\pi \times \int_{1}^{4} \left(4x^{2}+28x+49-\frac{6^{2}}{x^{2}}\right) dx\) - step2: Evaluate the integral: \(\pi \times 414\) - step3: Reorder the terms: \(414\pi \) The volume of the solid of revolution obtained by rotating the region \( R \) about the \( x \)-axis is \( 414\pi \) cubic units.

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The Deep Dive

To find the volume of the solid of revolution obtained by rotating the region \( R \) around the \( x \)-axis, we can use the method of washers. The volume \( V \) is given by the integral \[ V = \pi \int_{a}^{b} \left( f(x)^2 - g(x)^2 \right) \, dx \] In our case, \( f(x) = 2x + 7 \) and \( g(x) = \frac{6}{x} \), with the limits of integration being from \( x = 1 \) to \( x = 4 \). Let's calculate the volume step by step: 1. Compute \( f(x)^2 \) and \( g(x)^2 \): \[ f(x)^2 = (2x + 7)^2 = 4x^2 + 28x + 49 \] \[ g(x)^2 = \left(\frac{6}{x}\right)^2 = \frac{36}{x^2} \] 2. Set up the integral: \[ V = \pi \int_{1}^{4} \left( (4x^2 + 28x + 49) - \frac{36}{x^2} \right) \, dx \] 3. Combine the terms: \[ V = \pi \int_{1}^{4} \left( 4x^2 + 28x + 49 - \frac{36}{x^2} \right) \, dx \] 4. Evaluate the integral. We compute it term by term: \[ \int 4x^2 \, dx = \frac{4}{3}x^3, \quad \int 28x \, dx = 14x^2, \quad \int 49 \, dx = 49x, \quad \int \frac{36}{x^2} \, dx = -\frac{36}{x} \] 5. Now substituting the limits \( x = 1 \) and \( x = 4 \): \[ V = \pi \left[ \left( \frac{4}{3}(4^3) + 14(4^2) + 49(4) + \frac{36}{1} \right) - \left( \frac{4}{3}(1^3) + 14(1^2) + 49(1) + \frac{36}{1} \right) \right] \] 6. Calculate: \[ \frac{4}{3}(64) + 14(16) + 49(4) - \left( \frac{4}{3}(1) + 14(1) + 49(1) \right) = \frac{256}{3} + 224 + 196 - \left( \frac{4}{3} + 14 + 49 \right) \] Combine everything: \[ \frac{256}{3} + 224 + 196 = \frac{256}{3} + \frac{672}{3} + \frac{588}{3} = \frac{1516}{3} \] On the left for x=1: \[ \frac{4}{3} + 14 + 49 = \frac{4}{3} + 63 = \frac{193}{3} \] Now, \[ V = \pi \left( \frac{1516}{3} - \frac{193}{3} \right) = \pi \left( \frac{1323}{3} \right) = 441\pi \] Thus, the volume of the solid of revolution is \[ \boxed{441\pi} \]

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