Question

Find the volume by revolving the region bounded by the equations

f(x)=3-x² and y=1 Revolve around the x-Axis.

Answer 1

The volume of the solid formed by revolving the region bounded by f(x) = 3 -
`x^2` and y = 1 around the x-axis is 8
`\pi` cubic units.

Step-by-step explanation:

To find the volume of revolution, we use the disk method. The region enclosed by the curves (f(x) = 3 -
`x^2`) and y = 1 is a semicircle with a radius of 2 units (since (f(x) intersects y = 1 at (x =
`\pm 1`)). When rotated around the x-axis, it forms a solid shape. The formula for the volume of revolution in this case is (V =
`\pi \int_(a)^(b) [f(x)]^2 \,dx`), where
`([f(x)]^2\)` represents the function squared due to the disk method.

Here, the integral is evaluated from -1 to 1 (a = -1), b = 1), with (f(x) = 3 -
`x^2`). Therefore, (V =
`\pi \int_(-1)^(1) (3 - x^2)^2 \,dx`). After computing the integral, the resulting volume equals (8
`\pi`) cubic units.

This process involves slicing the area into infinitely thin disks perpendicular to the x-axis, summing their volumes, and integrating across the given interval. The (y = 1) line acts as the outer limit of the solid generated, with (f(x)) as the inner boundary. Integrating the squares of (f(x)) over this range provides the volume of the solid of revolution formed.

The disk method is a fundamental technique in calculus used for finding volumes of solids formed by revolving regions bounded by functions around an axis. In this case, the integration of the squared function over the specified interval yields the precise volume of the resulting solid.