The region \( R_2 \) is bounded by the lines \( y = 2 \sqrt[4]{x} \), \( y = 2 \), and the y-axis.
To find the volume generated by rotating \( R_2 \) about the line \( AB \) (which is the line \( y = 2 \)), we use the disk method. The radius of the disk at a point \( x \) is \( 2 - 2 \sqrt[4]{x} \).
The limits of integration for \( x \) are from \( 0 \) to \( 1 \) because \( y = 2 \sqrt[4]{x} \) intersects \( y = 2 \) at \( x = 1 \).
The volume \( V \) is given by:
\[ V = \pi \int_{0}^{1} \left(2 - 2 \sqrt[4]{x}\right)^2 \, dx \]
\[ \left(2 - 2 \sqrt[4]{x}\right)^2 = 4 - 8 \sqrt[4]{x} + 4 \sqrt[4]{x^2} \]
Since \( \sqrt[4]{x^2} = \sqrt{x} \), the integrand becomes:
\[ 4 - 8 \sqrt[4]{x} + 4 \sqrt{x} \]
\[ V = \pi \int_{0}^{1} \left(4 - 8 \sqrt[4]{x} + 4 \sqrt{x}\right) \, dx \]
\[ V = \pi \left[ 4x - \frac{8}{5} x^{5/4} + \frac{8}{3} x^{3/2} \right]_{0}^{1} \]
\[ V = \pi \left( 4(1) - \frac{8}{5}(1) + \frac{8}{3}(1) - (0 - 0 + 0) \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3} \right) \]
\[ V = \pi \left( 4 - \frac{8}{5} + \frac{8}{3}
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