Answer:
[tex]\mathbf{ \dfrac{128}{3}}[/tex]
Step-by-step explanation:
Given that:
[tex]y^2 + z^2 = 16; \\ \\ where; \ x = 2y , x =0, z= 0[/tex]
[tex]Replace \ z = 0 \ \text{in the given equation ;} y^2+z^2=16[/tex]
[tex]Then;[/tex]
[tex]y^2 = 16 \\ \\ y = \sqrt{16} \\ \\ y = \pm 4[/tex]
[tex]\text{So; in 1st octant \ limits of y }= 0 \to 4 \ \text{and for x }= 0 \to 2y[/tex]
∴
[tex]\text{Volume of solid: (V) = }\int ^{4}_{y=0} \int ^{2y}_{x=0} \sqrt{16-y^2} \ dxdy \\ \\ = \int^4_{y=0} \sqrt{16 -y^2} \ (x)^{2y} _o \ dy \\ \\ = \int^4_{y=0}\sqrt{16 -y^2} (2y -0) \ dy \\ \\ V = \int^{4}_{y=0} \ 2y \sqrt{16 - y^2} \ dy[/tex]
[tex]Now, let:\\\\ 16 - y^2 = u \\ \\ -2ydy = du \\ \\ 2ydy = -du \\ \\ Hence, when \ y = 0; u = 16 \\ \\ and \\ \\ when \ y = 4 \ then \ u = 0[/tex]
∴
[tex]V = \int ^4 _{y=0} \ 2y \sqrt{16 -y^2 } \ dy \\ \\ = \int ^0_{16} - \sqrt{u}\ du \\ \\ = - \int^0_{16} \ u^{^{\dfrac{1}{2}}} \ du \\ \\ = \Bigg [\dfrac{u^{^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1 } \Bigg]^0_{16} \\ \\ =\Bigg[ -\dfrac{u^{\dfrac{3}{2}}}{\dfrac{3}{2}}^0_{16} \Bigg] \\ \\[/tex]
[tex]= - \dfrac{2}{3}\Big( u^{\dfrac{3}{2}}\Big)^0_{16} \\ \\ = - \dfrac{2}{3} \Big(0^{^\dfrac{3}{2}}}- 16^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(0 - (4^2)^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(-(4)^{3}}\Big )[/tex]
[tex]= - \dfrac{2}{3} \Big(64\Big )[/tex]
∴
[tex]\mathbf{V = \dfrac{128}{3}}[/tex]
