Questions: Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis. y=36-x^2 y=0 x=4 x=6

Solution Steps

To find the volume of the solid generated by revolving the region bounded by the given equations about the y-axis, we can use the method of cylindrical shells. The volume \( V \) of the solid of revolution is given by the integral:

\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]

where \( f(x) = 36 - x^2 \), and the bounds \( a \) and \( b \) are 4 and 6, respectively.

We are given the region bounded by the equations: \[ \begin{array}{l} y = 36 - x^2 \\ y = 0 \\ x = 4 \\ x = 6 \end{array} \] We need to find the volume of the solid generated by revolving this region about the \( y \)-axis.

To find the volume using the method of cylindrical shells, we use the formula: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] where \( f(x) = 36 - x^2 \), and the bounds \( a \) and \( b \) are 4 and 6, respectively.

The integral to be evaluated is: \[ V = 2\pi \int_{4}^{6} x (36 - x^2) \, dx \] Evaluating this integral, we get: \[ V = 2\pi \left[ \int_{4}^{6} 36x \, dx - \int_{4}^{6} x^3 \, dx \right] \] \[ V = 2\pi \left[ 18x^2 \Big|_{4}^{6} - \frac{x^4}{4} \Big|_{4}^{6} \right] \] \[ V = 2\pi \left[ 18(36) - 18(16) - \frac{6^4}{4} + \frac{4^4}{4} \right] \] \[ V = 2\pi \left[ 648 - 288 - 324 + 64 \right] \] \[ V = 2\pi \left[ 100 \right] \] \[ V = 200\pi \]

Converting \( 200\pi \) to decimal form, we get: \[ V \approx 628.3185 \]

Final Answer