# How do you find the volume of the solid bounded by the coordinate planes and the plane #2x+y+z=3#?

Here is a reference to a similar problem that I will mimic.

The base of the solid is found my setting

From the above graph it is easy to see that x goes from 0 to 1.5 and y goes from 0 to

The height of the solid is

This allows us to write the following equation:

I used WolframAlpha to evaluate the integral:

V = 2.25

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Volume =

The coordinate planes are given by

The given tetrahedron is a solid that lies above the triangle

The line joining

# y-0 = ((3-0)/(0-3/2)) (x-3/2) #

# :. y = (-2)(x-3/2) #

# :. y = -2x+3 #

And so the region

# R = {(x, y) | 0 le x le 3/2, 0 le y le -2x+3 } #

And the volume,

integral of the function

# :. V = int int_R (3-2x-y) \ dA#

# " "= int_0^(3/2) int_0^(-2x+3) (3-2x-y) \ dy \ dx#

As with all multiple integrals we work from the inside out, and we integrate wrt the appropriate variable, whilst treating other variables as constant so let's deal with the inner iterated integral first, and we get:

# int_0^(-2x+3) (3-2x-y) \ dy = [3y-2xy-1/2y^2]_(y=0)^(y=-2x+3)#

# " "= 3(-2x+3)-2x(-2x+3)-1/2(-2x+3)^2 - 0#

# " "= -6x+9+4x^2-6x-1/2(4x^2-12x+9)#

# " "= -12x+9+4x^2-2x^2+6x-9/2#

# " "= 2x^2-6x+9/2#

And so our double integral becomes:

# V = int_0^(3/2) 2x^2-6x+9/2 \ dx#

# \ \ \ = [ 2/3x^3-3x^2+9/2x ]_0^(3/2)#

# \ \ \ = 2/3(27/8)-3(9/4)+9/2(3/2) - 0#

# \ \ \ = 9/4-27/4+27/4#

# \ \ \ = 9/4#

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A "pyramid" could be defined as the result of joining each point on any plane shape (such as a square, circle, or any polygon) to a point (the vertex) outside the plane, and its perpendicular height would be the shortest distance from the plane to that vertex.

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An yet another approach:

So we can define the tetrahedron vertices by the vectors

The the volume of the tetrahedron is given by:

We can evaluate this very quickly using the determinant with the vertex vectors as rows or columns):

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To find the volume of the solid bounded by the coordinate planes and the plane (2x+y+z=3), you would typically use triple integration.

You integrate over the region bounded by the coordinate planes and the given plane. The integral represents the volume of the solid within that region.

The integral would be set up as follows:

[ \iiint_{D} dV ]

Where (D) represents the region bounded by the coordinate planes and the plane (2x+y+z=3). You would then integrate (1) over this region with respect to (x), (y), and (z).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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