Question

find the volume of the solid whose base is the region in the xy-plane that is bounded py the parabola (y = 4 – x^2) and the line y = 3x, while the top of the solid is ounded by the plate z = x + 4.

Answer 1

The volume of the solid bounded by the region in the xy-plane, the parabola
`\(y = 4 - x^2\)`, the line
`\(y = 3x\)`, and the plane
`\(z = x + 4\)` is
`\((64)/(3)\)` cubic units.

Step-by-step explanation:

To find the volume of the solid, first determine the limits of integration by finding the points of intersection between the parabola and the line. Set
`\(y = 4 - x^2\) equal to \(y = 3x\)`to find the intersection points:

`\(4 - x^2 = 3x\)`

`\(x^2 + 3x - 4 = 0\)`

`\((x + 4)(x - 1) = 0\)`

`\(x = -4\) or \(x = 1\)`

The limits of integration for x are from -4 to 1. Now, set up the integral to find the volume using the formula for volume of a solid of revolution:

`\[V = \int_(-4)^(1) A(x) \, dx\]`

The area of the cross-section at a given x-value is the difference between the functions:
`\(A(x) = (4 - x^2) - 3x\)`. Thus, the integral becomes:

`\[V = \int_(-4)^(1) ((4 - x^2) - 3x) \, dx\]`

Solve the integral:

`\[V = \int_(-4)^(1) (4 - x^2 - 3x) \, dx\]`

`\[V = \left[4x - (x^3)/(3) - (3x^2)/(2)\right]_(-4)^(1)\]`

`\[V = \left[(4 - (1)/(3) - (3)/(2)) - (-16 + (64)/(3) + 24)\right]\]`

`\[V = (64)/(3)\]`

Therefore, the volume of the solid is
`\((64)/(3)\)` cubic units.