Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis. y = 36 - x², y = 0, x = 4, x = 6?

Answer 1

The volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is
`\[V = (1152)/(5)\pi\]` cubic units.

Step-by-step explanation:

To find the volume of the solid formed by revolving the region enclosed by the given equations about the y-axis, we can use the disk method. The region is bounded by the curves
`\(y = 36 - x^2\)`, (y = 0), (x = 4), and (x = 6). First, identify the points of intersection by setting
`\(y = 36 - x^2\)` equal to (0), which gives us
`\(x = \pm 6\)`. The limits of integration for the disk method will be from (x = -6) to (x = 6).

Next, the volume of an infinitesimally thin disk at a given x-value is calculated using the formula
`\(V_{\text{disk}} = \pi y^2\Delta x\)`, where (y) is the distance from the axis of rotation to the curve. In this case,
`\(y = 36 - x^2\)`. Integrating this expression from (x = -6) to (x = 6) gives us the total volume of the solid, which is
`\[V = \int_(-6)^(6) \pi (36 - x^2)^2 \,dx\]`.

Evaluating this integral leads to the final answer
`\[V = (1152)/(5)\pi\]`cubic units. This result represents the volume of the solid generated by revolving the specified region about the y-axis, providing a comprehensive understanding of the three-dimensional space enclosed by the given functions.