Questions: Find the volume of the solid generated by revolving the shaded region about the y-axis The volume is cubic units. (Type an exact answer, using π as needed.)

Solution Steps

The given function is \( x = 5 \tan(y) \). The limits for \( y \) are from \( 0 \) to \( \frac{\pi}{6} \).

To find the volume of the solid generated by revolving the region about the y-axis, we use the formula for the volume of revolution: \[ V = \pi \int_{a}^{b} [f(y)]^2 \, dy \] Here, \( f(y) = 5 \tan(y) \), \( a = 0 \), and \( b = \frac{\pi}{6} \).

Substitute \( f(y) \) into the integral: \[ V = \pi \int_{0}^{\frac{\pi}{6}} (5 \tan(y))^2 \, dy \] \[ V = 25\pi \int_{0}^{\frac{\pi}{6}} \tan^2(y) \, dy \]

Use the identity \( \tan^2(y) = \sec^2(y) - 1 \): \[ V = 25\pi \int_{0}^{\frac{\pi}{6}} (\sec^2(y) - 1) \, dy \] \[ V = 25\pi \left[ \int_{0}^{\frac{\pi}{6}} \sec^2(y) \, dy - \int_{0}^{\frac{\pi}{6}} 1 \, dy \right] \]

Evaluate the integrals: \[ \int_{0}^{\frac{\pi}{6}} \sec^2(y) \, dy = \tan(y) \Big|_{0}^{\frac{\pi}{6}} = \tan\left(\frac{\pi}{6}\right) - \tan(0) = \frac{1}{\sqrt{3}} - 0 = \frac{1}{\sqrt{3}} \] \[ \int_{0}^{\frac{\pi}{6}} 1 \, dy = y \Big|_{0}^{\frac{\pi}{6}} = \frac{\pi}{6} - 0 = \frac{\pi}{6} \]

Combine the results: \[ V = 25\pi \left( \frac{1}{\sqrt{3}} - \frac{\pi}{6} \right) \]

Final Answer