Question

Find the volume of the region bounded by z=12−y,z=y,y=x2, and y=6−x2. (Use symbolic notation and fractions where needed.)

Answer 1

The volume V of the region = 80

Explanation:

The objective is to find the volume of the solid region bounded by z = 12 - y

Given that the portion of solid between the planes z = 12 - y and z = y; over the region R in the xy - plane bounded by y = x² and y = 6 - x²

Thus; R = { (x,y)/-2 ≤ x ≤ 2, x² ≤ y 6 - x² }

Then the volume is computed as follows:

`V = \int \int_R [(12 -y) -y] \ dy. dx`

`V = \int^2_(-2) \int ^(6-x^2)_(x^2) (12-2y) \ dy \ dx`

`V = \int^2_(-2)[ (12y-y^2]^(6-x^2)_(x^2) \ dx`

`V = \int^2_(-2) \begin {bmatrix}{ 12 [6-x^2-x^2] - [(6-x^2)^2-(x^2)^2] \end {bmatrix} \ dx`

`V = \int^2_(-2) \begin {bmatrix}{ 12 [6-2x^2] - [36-12x^2]] \end {bmatrix} \ dx`

`V = \int^2_(-2) \begin {bmatrix}{ 72 -24x^2-36+12x^2 \end {bmatrix} \ dx`

`V = \int^2_(-2) [36-12x ^2] \ dx`

`V = \begin {bmatrix} 36x - (12x^3)/(3)\end {bmatrix}^2_(-2)`

`V =2 \begin {bmatrix} 72 - (96)/(3)\end {bmatrix}`

V = 80