Question

Find the volume of the given solid. Bounded by the cylinder y2 + z2 = 16 and the planes x = 2y, x = 0, z = 0 in the first octant

Answer 1

Answer:

$\mathbf{ \dfrac{128}{3}}$

Step-by-step explanation:

Given that:

$y^2 + z^2 = 16; \\ \\ where; \ x = 2y , x =0, z= 0$

$Replace \ z = 0 \ \text{in the given equation ;} y^2+z^2=16$

$Then;$

$y^2 = 16 \\ \\ y = \sqrt{16} \\ \\ y = \pm 4$

$\text{So; in 1st octant \ limits of y }= 0 \to 4 \ \text{and for x }= 0 \to 2y$

∴

$\text{Volume of solid: (V) = }\int ^{4}_{y=0} \int ^{2y}_{x=0} \sqrt{16-y^2} \ dxdy \\ \\ = \int^4_{y=0} \sqrt{16 -y^2} \ (x)^{2y} _o \ dy \\ \\ = \int^4_{y=0}\sqrt{16 -y^2} (2y -0) \ dy \\ \\ V = \int^{4}_{y=0} \ 2y \sqrt{16 - y^2} \ dy$

$Now, let:\\\\ 16 - y^2 = u \\ \\ -2ydy = du \\ \\ 2ydy = -du \\ \\ Hence, when \ y = 0; u = 16 \\ \\ and \\ \\ when \ y = 4 \ then \ u = 0$

∴

$V = \int ^4 _{y=0} \ 2y \sqrt{16 -y^2 } \ dy \\ \\ = \int ^0_{16} - \sqrt{u}\ du \\ \\ = - \int^0_{16} \ u^{^{\dfrac{1}{2}}} \ du \\ \\ = \Bigg [\dfrac{u^{^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1 } \Bigg]^0_{16} \\ \\ =\Bigg[ -\dfrac{u^{\dfrac{3}{2}}}{\dfrac{3}{2}}^0_{16} \Bigg]$

$= - \dfrac{2}{3}\Big( u^{\dfrac{3}{2}}\Big)^0_{16} \\ \\ = - \dfrac{2}{3} \Big(0^{^\dfrac{3}{2}}}- 16^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(0 - (4^2)^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(-(4)^{3}}\Big )$

$= - \dfrac{2}{3} \Big(64\Big )$

∴

$\mathbf{V = \dfrac{128}{3}}$