Use a triple integral to find the volume of the tetrahedron T bounded by the planes x+2y+z=2, x=2y, x=0 and z=0

Answer:

Volume of the Tetrahedron T =[tex]\frac{1}{3}[/tex]

Step-by-step explanation:

As given, The tetrahedron T is bounded by the planes x + 2y + z = 2, x = 2y, x = 0, and z = 0

We have,

z = 0 and x + 2y + z = 2

⇒ z = 2 - x - 2y

∴ The limits of z are :

0 ≤ z ≤ 2 - x - 2y

Now, in the xy- plane , the equations becomes

x + 2y = 2 , x = 2y , x = 0 ( As in xy- plane , z = 0)

Firstly , we find the intersection between the lines x = 2y and x + 2y = 2

∴ we get

2y + 2y = 2

⇒4y = 2

⇒y = [tex]\frac{2}{4} = \frac{1}{2}[/tex] = 0.5

⇒x = 2([tex]\frac{1}{2}[/tex]) = 1

So, the intersection point is ( 1, 0.5)

As we have x = 0 and x = 1

∴ The limits of x are :

0 ≤ x ≤ 1

Also,

x = 2y

⇒y = [tex]\frac{x}{2}[/tex]

and x + 2y = 2

⇒2y = 2 - x

⇒y = 1 - [tex]\frac{x}{2}[/tex]

∴ The limits of y are :

[tex]\frac{x}{2}[/tex] ≤ y ≤ 1 - [tex]\frac{x}{2}[/tex]

So, we get

Volume = [tex]\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}\int\limits^{2-x-2y}_{z=0} {dz} \, dy \, dx[/tex]

             = [tex]\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{[z]}\limits^{2-x-2y}_0 {} \, \, dy \, dx[/tex]

             = [tex]\int\limits^1_0 {\int\limits^{1-\frac{x}{2}}_{y = \frac{x}{2}}{(2-x-2y)} \, \, dy \, dx[/tex]

             = [tex]\int\limits^1_0 {[2y-xy-y^{2} ]}\limits^{1-\frac{x}{2}} _{\frac{x}{2} } {} \, \, dx[/tex]

             = [tex]\int\limits^1_0 {[2(1-\frac{x}{2} - \frac{x}{2}) -x(1-\frac{x}{2} - \frac{x}{2}) -(1-\frac{x}{2}) ^{2} + (\frac{x}{2} )^{2} ] {} \, \, dx[/tex]

             = [tex]\int\limits^1_0 {(1 - 2x + x^{2} )} \, \, dx[/tex]

             = [tex]{(x - x^{2} + \frac{x^{3}}{3} )}\limits^1_0[/tex]

             = 1 - 1² + [tex]\frac{1^{3} }{3}[/tex] - 0 + 0 - 0

             = 1 - 1 + [tex]\frac{1 }{3}[/tex] =  [tex]\frac{1}{3}[/tex]

So, we get

Volume =[tex]\frac{1}{3}[/tex]

The volume of Tetrahedron will be [tex]\frac{1}{3}[/tex] units.

Given,

The tetrahedron T is bounded by the planes,

[tex]x+2y+z=2......(1)\\ x=2y....(2)\\ x=0......(3) \\ z=0.....(4)[/tex]

From equation (1),

[tex]z=2-x-2y[/tex]

So the limits of z will be from 0 to [tex]2-x-2y[/tex].

Now, from equation (2),

[tex]y=\frac{x}{2}[/tex]

and from equation (1), putting z=0 we get,

[tex]x+2y=2\\y=1-\frac{x}{2}[/tex]

So the limits of y will be from [tex]\frac{x}{2}[/tex] to [tex]1-\frac{x}{2}[/tex].

On solving equation (1), for x we get

[tex]x+x+0=2[/tex] [tex](x=2y \ and \ z=0 )[/tex]

[tex]x=1[/tex].

So the limits of x will be from 0 to 1.

The volume of tetrahedron will be,

[tex]V=\int\limits^1_0 \, dx \int\limits^{1-\frac{x}{2} }_{\frac{x}{2} } \, dy \int\limits^{2-x-2y}_0 \, dz[/tex]

[tex]V=\int\limits^1_0 \, dx \int\limits^{1-\frac{x}{2} }_{\frac{x}{2} } \, dy [2-x-2y-0][/tex]

[tex]V=\int\limits^1_0 \, dx (1-2x+x^2)\\[/tex]

[tex]V=1-1+\frac{1}{3}[/tex]

[tex]V=\dfrac{1}{3}[/tex]

Hence the volume of tetrahedron is [tex]\frac{1}{3}[/tex] units.

For more details on Triple integration follow the link:

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