Question

Find the volume in the first octant bounded by x+y+z=9, and the inside cylinder 3y=27-x^3

Answer 1

$\text{setting } z=0, x+y=9 \\ \text{solving } \{x+y=9,3y=27-x^3\} \text{simultaneously,} \\ y=9-x\implies3(9-x)=27-x^3\implies x(x^2-3)=0 \\ \implies x=-\sqrt{3},0,\sqrt{3} \\ \text{The volume is given as } V=\int\limits_{ - \sqrt 3 }^{\sqrt 3 } {\int_{9 - \frac{{{x^3}}}{3}}^{9 - x} {\int_0^{9 - x - y} {dzdydx} } } \\ =\int_{ - \sqrt 3 }^{\sqrt 3 } {\int_{9 - \frac{{{x^3}}}{3}}^{9 - x} {\left( {9 - x - y} \right)} dydx} =\int_{ - \sqrt 3 }^{\sqrt 3 } {\left[ {9y - xy - \frac{{{y^2}}}{2}} \right]_{9 - \frac{{{x^3}}}{3}}^{9 - x}} dx \\ = \int_{ - \sqrt 3 }^{\sqrt 3 } {\left( {\frac{1}{2}{x^2} - \frac{1}{3}{x^4} + \frac{1}{{18}}{x^6}} \right)} dx = \left[ {\frac{1}{6}{x^3} - \frac{1}{{15}}{x^5} + \frac{1}{{126}}{x^7}} \right]_{ - \sqrt 3 }^{\sqrt 3 } \\ = \frac{{8\sqrt 3 }}{{35}}$