Question

Find the volume of the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through (2 comma 0 comma 0 )​, (0 comma 5 comma 0 )​, and (0 comma 0 comma 4 ).

Answer 1

Answer:

The volume of the tetrahedron is:

$\frac{50}{3}=16.667$

Step-by-step explanation:

The volume of the tetrahedron is given by the intersection of the planes x = 0, y = 0, z = 0 and the plane formed by the three points given.

The equation of the plane formed by the three points is:

Points: (2,0,0);(0,5,0);(0,0,4)

$\pi :\frac{x}{2} +\frac{y}{5} +\frac{z}{4} =1$

It can also be expressed as:

$10x + 4y+5z=20$

We have to calculate the triple integral, therefore we must define the domain:

The values of x are given by:

0≤x≤2

We will integrate the values of y between the y = 0 axis and the line formed when z = 0:

z=0 ⇒ 10x + 4y=2 ⇒ $y=\frac{20-10x}{4} =$

$0\leq y \leq 5-\frac{5}{2}x$

We will integrate the values of z between the plane z = 0 and the plane 10x + 4y+5z=20

10x + 4y+5z=20 ⇒ $z=\frac{20-10x-4y}{5} =4-2x-\frac{4}{5}y$

$0\leq z \leq 4-2x-\frac{4}{5}y$

The volume of the tetrahedron is:

$\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} \ \ \ \int\limits^{4-2x-\frac{4}{5}y }_{0} {z} \, dz\, dy \,dx$

$\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} \frac{z^2}{2}|^{4-2x-\frac{4}{5}y}_0\, dy \,dx$

$\frac{z^2}{2}|^{z=4-2x-\frac{4}{5}y}_{z=0}=\frac{5}{25}(5x+2(y-5))^2$

$\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} {\frac{5}{25}(5x+2(y-5))^2}\, dy \,dx$

$\int\limits^{2}_{0} {-\frac{25}{6}(x-2)^3} dx=-\frac{25}{24} (x-2)^4|^{2}_{0}=\frac{50}{3}$