Question

Use a triple integral to find the volume of the given solid. The tetrahedron enclosed by the coordinate planes and the plane 2x + y + z = 2

Answer 1

Answer:

$\mathbf{V = \dfrac{2}{3}}$

Step-by-step explanation:

Consider the tetrahedron enclosed by the coordinate planes and the plane 2x + y + z = 2

Let A be the region obtained by projecting the volume(V) onto the xy-plane.

Similarly, the plane 2x + y + z = 2 intersects with the xy-plane in y = 2 - 2x.

Using the vertical strips, the region A is to the xy-plane can be expressed as:

$A = \{(x,y) | 0 \le x \le 1, 0 \le y \le 2 - 2x\}$

Thus, the volume of the solid can be calculated as follows;

$V = \int^1_0 \int ^{2-2x}_{0} \int ^{2-2x-y}_{0} \ dz \ dy \ dx$

$V = \int^1_0 \int ^{2-2x}_{0} (2-2x-y) \ dy \ dx$

$V = \int^1_0 \bigg [2\times y -2x\times y - \dfrac{y^2}{2} \bigg]^{2-2x}_{0} \ dx$

$V = \int^1_0 \bigg [2\times (2-2x) -2x\times (2-2x) - \dfrac{(2-2x)^2}{2} \bigg]\ dx$

$V = \int^1_0 \bigg [2x^2-4x+2 \bigg]\ dx$

$V = \bigg [2\dfrac{x^3}{3}-\dfrac{4x^2}{2}+2 x\bigg]^1_0$

$V = \bigg [2\dfrac{(1)^3}{3}-\dfrac{4(1)^2}{2}+2 (1)-0\bigg]$

$V = \dfrac{2}{3}- 2 + 2$

$\mathbf{V = \dfrac{2}{3}}$