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14. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines. \[ y=\sqrt{x}, y=0, x=4 \] a. Rotated around the \( x \)-axis b. Rotated around the \( y \)-axis c. Rotated around the line \( x=4 \) d. Rotated around the line \( x=6 \)

Ask by Kirk Fitzgerald. in the United States
Dec 17,2024

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The volumes for the different rotations are as follows: a. Rotated around the \( x \)-axis: \( 8\pi \) b. Rotated around the \( y \)-axis: \( 8\pi \) c. Rotated around the line \( x = 4 \): \( 8\pi \) d. Rotated around the line \( x = 6 \): \( 16\pi \)

Solución

To find the volume of the solid generated by revolving the region bounded by the graphs of the equations \( y = \sqrt{x} \), \( y = 0 \), and \( x = 4 \) around the specified axes, we will use the method of disks/washers for revolutions around the \( x \)-axis and \( y \)-axis, and the shell method for revolutions around vertical lines. ### a. Rotated around the \( x \)-axis When rotating around the \( x \)-axis, we use the disk method. The volume \( V \) is given by: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] Here, \( f(x) = \sqrt{x} \), and the bounds are from \( x = 0 \) to \( x = 4 \): \[ V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx = \pi \int_{0}^{4} x \, dx \] Calculating the integral: \[ V = \pi \left[ \frac{x^2}{2} \right]_{0}^{4} = \pi \left[ \frac{4^2}{2} - \frac{0^2}{2} \right] = \pi \left[ \frac{16}{2} \right] = 8\pi \] ### b. Rotated around the \( y \)-axis When rotating around the \( y \)-axis, we use the shell method. The volume \( V \) is given by: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] Here, \( f(x) = \sqrt{x} \), and we need to express \( x \) in terms of \( y \). From \( y = \sqrt{x} \), we have \( x = y^2 \). The bounds for \( y \) are from \( y = 0 \) to \( y = \sqrt{4} = 2 \): \[ V = 2\pi \int_{0}^{2} (y^2) \cdot y \, dy = 2\pi \int_{0}^{2} y^3 \, dy \] Calculating the integral: \[ V = 2\pi \left[ \frac{y^4}{4} \right]_{0}^{2} = 2\pi \left[ \frac{2^4}{4} - 0 \right] = 2\pi \left[ \frac{16}{4} \right] = 2\pi \cdot 4 = 8\pi \] ### c. Rotated around the line \( x = 4 \) For rotation around the line \( x = 4 \), we again use the shell method. The volume \( V \) is given by: \[ V = 2\pi \int_{a}^{b} (R - r) \cdot h \, dy \] Where \( R \) is the distance from the line \( x = 4 \) to the curve, and \( r \) is the distance from the line to the axis of rotation. Here, \( R = 4 - y^2 \) and \( h = y \). The bounds for \( y \) are from \( 0 \) to \( 2 \): \[ V = 2\pi \int_{0}^{2} (4 - y^2) \cdot y \, dy = 2\pi \int_{0}^{2} (4y - y^3) \, dy \] Calculating the integral: \[ V = 2\pi \left[ 2y^2 - \frac{y^4}{4} \right]_{0}^{2} = 2\pi \left[ 2(2^2) - \frac{(2^4)}{4} \right] = 2\pi \left[ 8 - 4 \right] = 2\pi \cdot 4 = 8\pi \] ### d. Rotated around the line \( x = 6 \) For rotation around the line \( x = 6 \), we again use the shell method. The volume \( V \) is given by: \[ V = 2\pi \int_{a}^{b} (R - r) \cdot h \, dy \] Where \( R = 6 - y^2 \) and \( h = y \). The bounds for \( y \) are from \( 0 \) to \( 2 \): \[ V = 2\pi \int_{0}^{2} (6 - y^2) \cdot y \, dy = 2\pi \int_{0}^{2} (6y - y^3) \, dy \] Calculating the integral: \[ V = 2\pi \left[ 3y^2 - \frac{y^4}{4} \right]_{0}^{2} = 2\pi \left[ 3(2^2) - \frac{(2^4)}{4} \right] = 2\pi \left[ 12 - 4 \right] = 2\pi \cdot 8 = 16\pi \] ### Summary of Volumes - a. Rotated around the \( x \)-axis: \( 8\pi \) - b. Rotated around the \( y \)-axis: \( 8\pi \) - c. Rotated around the line \( x = 4 \): \( 8\pi \) - d. Rotated around the line \( x = 6 \): \( 16\pi \)

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The Deep Dive

To find the volume of the solid generated by revolving the region bounded by the curves about the specified lines, we can use the method of disks and shells. First, let's identify the area we are working with. The region is bounded by the curves \(y = \sqrt{x}\), \(y = 0\) (the x-axis), and the vertical line \(x = 4\). This forms a region that looks like a "slice" of a curve. ### a. Rotated around the \(x\)-axis When rotated around the \(x\)-axis, we can use the disk method. The volume \(V\) is given by: \[ V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx = \pi \int_{0}^{4} x \, dx = \pi \left[ \frac{x^2}{2} \right]_{0}^{4} = \pi \left( \frac{16}{2} - 0 \right) = 8\pi \] ### b. Rotated around the \(y\)-axis For rotation about the \(y\)-axis, we will use the shell method. The volume \(V\) is given by: \[ V = 2\pi \int_{0}^{4} x \cdot \sqrt{x} \, dx = 2\pi \int_{0}^{4} x^{3/2} \, dx = 2\pi \left[ \frac{x^{5/2}}{5/2} \right]_{0}^{4} = 2\pi \cdot \frac{2}{5} \cdot (4^{5/2}) = 2\pi \cdot \frac{2}{5} \cdot 32 = \frac{128\pi}{5} \] ### c. Rotated around the line \(x=4\) Using the shell method again, the volume generated is calculated by adjusting the radius. The volume \(V\) now is: \[ V = 2\pi \int_{0}^{4} (4-x) \cdot \sqrt{x} \, dx = 2\pi \int_{0}^{4} (4\sqrt{x} - x^{3/2}) \, dx \] Calculating each integral, we find: \[ = 2\pi \left[ \frac{8}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right]_{0}^{4} = 2\pi \left( \frac{8}{3} \cdot 8 - \frac{2}{5} \cdot 32 \right) = 2\pi \left( \frac{64}{3} - \frac{64}{5} \right) \] This gives a final volume of \(V = \frac{256\pi}{15}\). ### d. Rotated around the line \(x=6\) In this case, where we rotate about \(x=6\), we again use the shell method: \[ V = 2\pi \int_{0}^{4} (6-x) \cdot \sqrt{x} \, dx = 2\pi \int_{0}^{4} (6\sqrt{x} - x^{3/2}) \, dx \] Calculating the integrals gives: \[ = 2\pi \left[ 6 \cdot \frac{8}{3} - \frac{2}{5} \cdot 32 \right] = 2\pi \left( 16 - \frac{64}{5} \right) = 2\pi \left( \frac{80}{5} - \frac{64}{5} \right) = \frac{32\pi}{5} \] In summary, the volumes are: - **a.** \(8\pi\) - **b.** \(\frac{128\pi}{5}\) - **c.** \(\frac{256\pi}{15}\) - **d.** \(\frac{32\pi}{5}\)

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