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

Answer:

The volume V of the region = 80

Step-by-step 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:

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

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

$V = \int^2_{-2}[ (12y-y^2]^{6-x^2}_{x^2} \ dx$

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

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

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

$V = \int^2_{-2} [36-12x ^2] \ dx$

$V = \begin {bmatrix} 36x - \dfrac{12x^3}{3}\end {bmatrix}^2_{-2}$

$V =2 \begin {bmatrix} 72 - \dfrac{96}{3}\end {bmatrix}$

V = 80