Answer to Question #274497 in Calculus for nick

Question

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

Answer 1

setting $z=0, x+y=9$

solving $\left\{x+y=9,3 y=27-x^{3}\right\}$ simultaneously,

$\begin{aligned} &y=9-x \Longrightarrow 3(9-x)=27-x^{3} \Longrightarrow x\left(x^{2}-3\right)=0 \\ &\Longrightarrow x=-\sqrt{3}, 0, \sqrt{3} \end{aligned}$

$\begin{aligned} &\text { The volume is given as } V=\int_{-\sqrt{3}}^{\sqrt{3}} \int_{9-\frac{x^{3}}{3}}^{9-x} \int_{0}^{9-x-y} d z d y d x \\ &=\int_{-\sqrt{3}}^{\sqrt{3}} \int_{9-\frac{x^{3}}{3}}^{9-x}(9-x-y) d y d x=\int_{-\sqrt{3}}^{\sqrt{3}}\left[9 y-x y-\frac{y^{2}}{2}\right]_{9-\frac{z^{3}}{3}}^{9-x} d x \\ &=\int_{-\sqrt{3}}^{\sqrt{3}}\left(\frac{1}{2} x^{2}-\frac{1}{3} x^{4}+\frac{1}{18} x^{6}\right) d x=\left[\frac{1}{6} x^{3}-\frac{1}{15} x^{5}+\frac{1}{126} x^{7}\right]_{-\sqrt{3}}^{\sqrt{3}} \\ &=\frac{8 \sqrt{3}}{35} \end{aligned}$