Question

Find the volume of the region between the cylinder z=3y^2 and the xy-plane that is bounded by the planes x=0,x=1 ,y=-1 and . z = y2 x = 0 x = 1 y = − 1 y =1

Answer 1

The volume of the region V = 2

Explanation:

Given that:

`z_1 = 3y^2` ;

where initially;

`z_o = 0; \ x_o = 0; \ x_1 = 1; \ y_o= -1; \ y_1 = 1`

The volume of the region is given by a triple which is expressed as:

`V = \int_x \int_y \int_z \ dz \ dy \ dx`

`V = \int \limits ^(x_1 = 1)_(x_o=0) \int \limits ^(y_1 = 1)_(y_o=-1) \int \limits ^(z_1 = 3y^2)_(z_o=0) \ dz \ dy \ dx`

`V = \int \limits ^(1)_(0) \int \limits ^( 1)_(-1) \int \limits ^(3y^2)_(0) \ dz \ dy \ dx`

`V = \int \limits ^(1)_(0) \int \limits ^( 1)_(-1) \Bigg [z \Bigg]^(3y^2)_(0) \ dy \ dx`

`V = \int \limits ^(1)_(0) \int \limits ^( 1)_(-1) \Bigg [3y^2 \Bigg] \ dy \ dx`

`V = \int \limits ^(1)_(0) \Bigg [(3y^3)/(3) \Bigg]^1_(-1) \ dx`

`V = \int \limits ^(1)_(0) \Bigg [(3(1)^3)/(3)- (3(-1)^3)/(3) \Bigg] \ dx`

`V = \int \limits ^(1)_(0) \Bigg [1-(-1)\Bigg] \ dx`

`V =2 \Bigg [x \Bigg] ^1_0`

V = 2

Thus, the volume of the region is 2