Find the volume of the solid obtained by rotating the region bounded by the curves \[ 3 y x^{4}, y=1 \] about the line \( y=5 \)

To find the volume of the solid obtained by rotating the region bounded by the curves \( 3yx^4 \) and \( y=1 \) about the line \( y=5 \), we first need to determine the points of intersection of the curves. 1. **Finding the intersection points:** We set \( 3yx^4 = 1 \) to find the values of \( x \) where the curves intersect: \[ 3yx^4 = 1 \implies y = \frac{1}{3x^4} \] Setting \( y = 1 \): \[ 1 = \frac{1}{3x^4} \implies 3x^4 = 1 \implies x^4 = \frac{1}{3} \implies x = \left(\frac{1}{3}\right)^{1/4} = \frac{1}{\sqrt[4]{3}}. \] Thus, the intersection points occur at \( x = \pm \frac{1}{\sqrt[4]{3}} \). 2. **Setting up the volume integral:** We will use the washer method to find the volume of the solid of revolution. The volume \( V \) is given by the integral: \[ V = \pi \int_{a}^{b} \left( R^2 - r^2 \right) \, dx, \] where \( R \) is the outer radius and \( r \) is the inner radius. In this case, the outer radius \( R \) is the distance from the line \( y=5 \) to the curve \( y=1 \): \[ R = 5 - 1 = 4. \] The inner radius \( r \) is the distance from the line \( y=5 \) to the curve \( y = 3yx^4 \): \[ r = 5 - 3yx^4 = 5 - 3 \left( \frac{1}{3x^4} \right) x^4 = 5 - 1 = 4. \] However, we need to express \( y \) in terms of \( x \) for the inner radius: \[ y = 3yx^4 \implies y = \frac{1}{3x^4}. \] Thus, the inner radius becomes: \[ r = 5 - \frac{1}{3x^4}. \] 3. **Setting the limits of integration:** The limits of integration are from \( x = -\frac{1}{\sqrt[4]{3}} \) to \( x = \frac{1}{\sqrt[4]{3}} \). 4. **Calculating the volume:** The volume integral becomes: \[ V = \pi \int_{-\frac{1}{\sqrt[4]{3}}}^{\frac{1}{\sqrt[4]{3}}} \left( (5 - 1)^2 - \left( 5 - \frac{1}{3x^4} \right)^2 \right) \, dx. \] Simplifying the expression: \[ V = \pi \int_{-\frac{1}{\sqrt[4]{3}}}^{\frac{1}{\sqrt[4]{3}}} \left( 4^2 - \left( 5 - \frac{1}{3x^4} \right)^2 \right) \, dx. \] Expanding the square: \[ (5 - \frac{1}{3x^4})^2 = 25 - \frac{10}{3x^4} + \frac{1}{9x^8}. \] Thus, \[ V = \pi \int_{-\frac{1}{\sqrt[4]{3}}}^{\frac{1}{\sqrt[4]{3}}} \left( 16 - \left( 25 - \frac{10}{3x^4} + \frac{1}{9x^8} \right) \right) \, dx. \] Simplifying further: \[ V = \pi \int_{-\frac{1}{\sqrt[4]{3}}}^{\frac{1}{\sqrt[4]{3}}} \left( 16 - 25 + \frac{10}{3x^4} - \frac{1}{9x^8} \right) \, dx = \pi \int_{-\frac