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

Answer 1

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

The base of the solid is found my setting #z = 0#; this forms a right triangle with, the x axis as the base, the y axis as the height, and the line #y = 3-2x# as the hypotenuse.

From the above graph it is easy to see that x goes from 0 to 1.5 and y goes from 0 to #3 - 2x#:

The height of the solid is #z = 3 - y - 2x#

This allows us to write the following equation:

#V = int_0^1.5int_0^(3-2x)(3 - y - 2x)dydx#

I used WolframAlpha to evaluate the integral:

V = 2.25

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Answer 2

Volume = #9/4 \ "unit"^3#

The coordinate planes are given by #x = 0#, #y = 0# and #z = 0#. The volume is that of a tetrahedron whose vertices are the intersections of three of the four planes given. The intersection of #x = 0#, #y = 0# and #2x + y + z = 3# is #(0, 0, 3)#, Similarly, the other three vertices are #(3/2, 0, 0)#, #(0, 3, 0)# and the origin #(0, 0, 0)#.

The given tetrahedron is a solid that lies above the triangle #R# in the #xy#-plane that has vertices #(0, 0)#, #(3/2, 0)# and #(0, 3)#. So if we look at #xy#-plane we have:

The line joining #(0, 3)# and #(3/2, 0)# is given by:

# y-0 = ((3-0)/(0-3/2)) (x-3/2) #
# :. y = (-2)(x-3/2) #
# :. y = -2x+3 #

And so the region #R# is defined as:

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

And the volume, #V#, of the tetrahedron is the double
integral of the function #z=3-2x-y# over the region #R# as defined above.

# :. 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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Answer 3

#9/4#

The volume of a pyramid is one third the area of the base times the perpendicular height. The base is a right triangle in the #x-y# plane with vertices (0,0), (3/2,0), (0,3) so it has area #1/2 xx 3/2 xx 3=9/4# The volume is therefore one third of this, times the height #3# (being the distance from the #x-y# plane to the vertex #(0,0,3)# So the volume is #cancel(1/3) xx 9/4 xx cancel(3)#.

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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Answer 4

An yet another approach:

As in my earlier answer, The coordinate planes are given by #x = 0#, #y = 0# and #z = 0#. The volume is that of a tetrahedron whose vertices are the intersections of three of the four planes given. The intersection of #x = 0#, #y = 0# and #2x + y + z = 3# is #(0, 0, 3)#, Similarly, the other three vertices are #(3/2, 0, 0)#, #(0, 3, 0)# and the origin #(0, 0, 0)#.

So we can define the tetrahedron vertices by the vectors

# vec (OA)=<<(3/2, 0, 0)>> # # vec (OB)=<<(0, 0, 3)>> # # vec (OC)=<<(0, 3, 0)>> #

The the volume of the tetrahedron is given by:

# V=1/6*abs(vec (OA) *(vec(OB) xx vec(OC)))#
Where, #vec(OA) * (vec(OB) xx vec(OC))# is the scalar triple product

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

# V = 1/6 | (3/2,0,0), (0,3,0), (0,0,3) | # # \ \ = 1/6 (3/2)9 # # \ \ = 9/4 #
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Answer 5

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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Answer from HIX Tutor

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.

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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