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\begin{tabular}{l} Use the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \\ line \( x=5 \). \\ \( \qquad \begin{aligned} y=\sqrt{x} \\ y=0 \\ x=4\end{aligned} \) \\ \hline\end{tabular}

Ask by Johnson Bob. in the United States
Feb 25,2025

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The volume of the solid is \( \frac{416\pi}{15} \) cubic units.

Solución

We wish to find the volume of the solid formed by revolving the region bounded by   y = √x,  y = 0,  and  x = 4 about the vertical line x = 5 using the shell method. Step 1. Sketch the region. • The curve y = √x starts at (0,0) and rises to (4,2) since √4 = 2. • The line y = 0 is the x-axis. • The vertical line x = 4 is the right boundary. Thus, the region lies between x = 0 and x = 4 (with the left boundary implicitly at x = 0 because y = √x and y = 0 meet at x = 0). Step 2. Set up the shell method. Since we are revolving about the vertical line x = 5, it is natural to use vertical slices (dx). A typical slice at position x (where 0 ≤ x ≤ 4) has: • Height = √x (from y = 0 to y = √x). • Distance from the axis of rotation = radius = (5 − x). • Thickness = dx. The lateral surface area of the cylindrical shell is 2π(radius)(height) and its volume is 2π(5 − x)(√x) dx. Step 3. Write the integral for the volume. The volume V is given by   V = ∫[x=0 to 4] 2π (5 − x)(√x) dx     = 2π ∫₀⁴ (5√x − x√x) dx. Step 4. Evaluate the integral. First, note that x√x = x^(3/2). So,   V = 2π [ 5∫₀⁴ x^(1/2) dx − ∫₀⁴ x^(3/2) dx ]. Recall the power rule for integration:   ∫ x^n dx = x^(n+1)/(n+1). Thus,   ∫ x^(1/2) dx = (2/3)x^(3/2)  and  ∫ x^(3/2) dx = (2/5)x^(5/2). Substitute these into the integral:   V = 2π [ 5*(2/3)x^(3/2) − (2/5)x^(5/2) ] evaluated from 0 to 4,     = 2π [ (10/3)x^(3/2) − (2/5)x^(5/2) ] from x = 0 to x = 4. Now, evaluate at x = 4 (the lower limit gives 0): • 4^(3/2) = (4^(1))·(4^(1/2)) = 4·2 = 8. • 4^(5/2) = 4^(2)·(4^(1/2)) = 16·2 = 32. So,   V = 2π [ (10/3)*8 − (2/5)*32 ]     = 2π [ (80/3) − (64/5) ]. Find a common denominator for the terms inside the bracket:   (80/3) = (80×5)/(3×5) = 400/15  and  (64/5) = (64×3)/(5×3) = 192/15. Thus,   (80/3) − (64/5) = (400 − 192) / 15 = 208/15. Finally, multiply by 2π:   V = 2π (208/15) = (416π)/15. Step 5. Write the final answer. The volume of the solid is (416π)/15 cubic units.

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To find the volume of the solid generated by revolving the region bounded by the graphs of \( y = \sqrt{x} \), \( y = 0 \), and \( x = 4 \) about the line \( x = 5 \) using the shell method, we first need to identify the relevant dimensions and shape of the shell. The region we’re revolving is from \( x = 0 \) to \( x = 4 \), and the height of the shell at any point \( x \) is given by the function \( y = \sqrt{x} \). The formula for the volume \( V \) using the shell method is: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dx \] Here, the radius is the distance from the shell to the line \( x = 5 \), which is \( 5 - x \). The height is \( y = \sqrt{x} \). Thus, we have: \[ V = 2\pi \int_{0}^{4} (5 - x)(\sqrt{x}) \, dx \] Now, we will compute the integral: 1. Distribute \( (5 - x) \) with \( \sqrt{x} \): \[ V = 2\pi \int_{0}^{4} (5\sqrt{x} - x\sqrt{x}) \, dx \] or: \[ = 2\pi \int_{0}^{4} (5\sqrt{x} - x^{3/2}) \, dx \] 2. Evaluate the integral: - Integrate \( 5\sqrt{x} \): \[ \int 5\sqrt{x} \, dx = 5 \cdot \frac{2}{3} x^{3/2} = \frac{10}{3} x^{3/2} \] - Integrate \( x^{3/2} \): \[ \int x^{3/2} \, dx = \frac{2}{5} x^{5/2} \] 3. Combine and evaluate: \[ V = 2\pi \left[ \left( \frac{10}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right) \Bigg|_{0}^{4} \right] \] Evaluate at \( x = 4 \): \[ V = 2\pi \left[ \left( \frac{10}{3} (4)^{3/2} - \frac{2}{5} (4)^{5/2} \right) - 0 \right] \] \[ = 2\pi \left[ \frac{10}{3} (8) - \frac{2}{5} (32) \right] \] \[ = 2\pi \left[ \frac{80}{3} - \frac{64}{5} \right] \] Finding a common denominator (15): \[ = 2\pi \left[ \frac{400}{15} - \frac{192}{15} \right] \] \[ = 2\pi \left[ \frac{208}{15} \right] \] \[ = \frac{416\pi}{15} \] Thus, the volume \( V \) of the solid generated by revolving the given region around the line \( x = 5 \) is: \[ \boxed{\frac{416\pi}{15}} \]

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