Answer:
The volume of the region V = 2
Step-by-step explanation:
Given that:
[tex]z_1 = 3y^2[/tex] ;
where initially;
[tex]z_o = 0; \ x_o = 0; \ x_1 = 1; \ y_o= -1; \ y_1 = 1[/tex]
The volume of the region is given by a triple which is expressed as:
[tex]V = \int_x \int_y \int_z \ dz \ dy \ dx[/tex]
[tex]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[/tex]
[tex]V = \int \limits ^{1}_{0} \int \limits ^{ 1}_{-1} \int \limits ^{3y^2}_{0} \ dz \ dy \ dx[/tex]
[tex]V = \int \limits ^{1}_{0} \int \limits ^{ 1}_{-1} \Bigg [z \Bigg]^{3y^2}_{0} \ dy \ dx[/tex]
[tex]V = \int \limits ^{1}_{0} \int \limits ^{ 1}_{-1} \Bigg [3y^2 \Bigg] \ dy \ dx[/tex]
[tex]V = \int \limits ^{1}_{0} \Bigg [\dfrac{3y^3}{3} \Bigg]^1_{-1} \ dx[/tex]
[tex]V = \int \limits ^{1}_{0} \Bigg [\dfrac{3(1)^3}{3}- \dfrac{3(-1)^3}{3} \Bigg] \ dx[/tex]
[tex]V = \int \limits ^{1}_{0} \Bigg [1-(-1)\Bigg] \ dx[/tex]
[tex]V =2 \Bigg [x \Bigg] ^1_0[/tex]
V = 2
Thus, the volume of the region is 2
