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frez [133]
4 years ago
9

Can yall help me plsssssss

Mathematics
1 answer:
k0ka [10]4 years ago
7 0

Answer:

I do not know the anwser to that

Step-by-step explanation:

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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
Tanzania [10]

Answer:

Volume of the Tetrahedron T =\frac{1}{3}

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 = \frac{2}{4} = \frac{1}{2} = 0.5

⇒x = 2(\frac{1}{2}) = 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 = \frac{x}{2}

and x + 2y = 2

⇒2y = 2 - x

⇒y = 1 - \frac{x}{2}

∴ The limits of y are :

\frac{x}{2} ≤ y ≤ 1 - \frac{x}{2}

So, we get

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

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

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

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

             = \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

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

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

             = 1 - 1² + \frac{1^{3} }{3} - 0 + 0 - 0

             = 1 - 1 + \frac{1 }{3} =  \frac{1}{3}

So, we get

Volume =\frac{1}{3}

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Step-by-step explanation:

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