Question

Use a triple Integral to determine the volume of the tetrahedron with vertices (0,0,0), (3,0,0), (0,4,0) and (0,0,5).

Answer 1

The equation of the line that passes throughout the points (3,0,0) and (0,4,0) in the plane $xOy$ is $\frac{x}{3}+\frac{y}{4}=1$ and hence $y=4(1-\frac{x}{3})$. The equation of the plane that passes throughout the points (3,0,0), (0,4,0) and (0,0,5) is $\frac{x}{3}+\frac{y}{4}+\frac{z}{5}=1$ and thus $z=5(1-\frac{x}{3}-\frac{y}{4})$. It follows that the volume $V$ is uqual to

$V=\int_{0}^{3}dx\int_{0}^{4(1-\frac{x}{3})}dy\int_0^{5(1-\frac{x}{3}-\frac{y}{4})}dz= \int_{0}^{3}dx\int_{0}^{4(1-\frac{x}{3})}5(1-\frac{x}{3}-\frac{y}{4})dy= 5\int_{0}^{3}(4(1-\frac{x}{3})-4\frac{x}{3}(1-\frac{x}{3})-\frac{1}{8}(4(1-\frac{x}{3}))^2)dx= 5\int_{0}^{3}(4-\frac{4x}{3}-4\frac{x}{3}+\frac{4x^2}{9}-2+4\frac{x}{3}+2\frac{x^2}{9})dx= 5\int_{0}^{3}(2-\frac{4x}{3}+2\frac{x^2}{3})dx= 5(2x-\frac{2x^2}{3}+\frac{2x^3}{9})|_0^3=5(6-6+2)=10$