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

Answer:

The volume of the region V = 2

Step-by-step 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:

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

$V = \int \limits ^{x_1 = 1}_{x_o=0} \int \limits ^{y_1 = 1}_{y_o=-1} \int \limits ^{z_1 = 3y^2}_{z_o=0} \ dz \ dy \ dx$

$V = \int \limits ^{1}_{0} \int \limits ^{ 1}_{-1} \int \limits ^{3y^2}_{0} \ dz \ dy \ dx$

$V = \int \limits ^{1}_{0} \int \limits ^{ 1}_{-1} \Bigg [z \Bigg]^{3y^2}_{0} \ dy \ dx$

$V = \int \limits ^{1}_{0} \int \limits ^{ 1}_{-1} \Bigg [3y^2 \Bigg] \ dy \ dx$

$V = \int \limits ^{1}_{0} \Bigg [\dfrac{3y^3}{3} \Bigg]^1_{-1} \ dx$

$V = \int \limits ^{1}_{0} \Bigg [\dfrac{3(1)^3}{3}- \dfrac{3(-1)^3}{3} \Bigg] \ dx$

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

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

V = 2

Thus, the volume of the region is 2