Answer with explanation:
We have to find Volume of solid under the plane
7 x +2 y-z=0
And the plane is enclosed by the Parabolas
1. →y=x²
2.→x=y²
Equating (1) and (2)
[tex]\Rightarrow x=x^4\\\\\Rightarrow x^4-x=0\\\\\Rightarrow x (x^3-1)=0\\\\\Rightarrow x=0,1[/tex]
→x=0 , gives , y=0
→x=1 , gives , y=1
1.→ 0 ≤ x ≤ 1
2.→ 0 ≤ y ≤1
3.→ 0≤ z ≤ 7 x +2 y
Volume of required Region
[tex]=\int\int[\int\limits^{7 x+ 2 y}_0 {1} \, dz)] \, dA\\\\=\int\limits^1_0\int\limits^1_0 (z)\left \{ {{z=7 x+2 y} \atop {z=0}} \right \, dy \,d x \\\\=\int\limits^1_0\int\limits^1_0 (7 x+2 y) \, dy \,d x\\\\=\int\limits^1_0 [(7 x y+y^2)\left \{ {{y=1} \atop {y=0}} \right. ] d x\\\\=\int\limits^1_0 [7 x +1] \, dx\\\\=\frac{7 x^2}{2}+x\left \{ {{x=1} \atop {x=0}} \right\\\\=\frac{7}{2}+1\\\\=\frac{9}{2}[/tex]
Volume of the bounded region
[tex]=\frac{9}{2} (\text{Units})^3[/tex]
