For this plane, since it intersects with the #xy#, #xz#, and also #yz# planes, it renders one-fourth that a rhomboid pyramid. So, every we have to do is:

Find the intersectionsDetermine the length of every diagonal distanceFind the volume the the entire hypothetical rhomboid pyramidDivide through #4#The intersections room at the #x#, #y#, and #z# axis.

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So, the 3 intersections space #(2,0,0)#, #(0,4,0)#, and #(0,0,4)#, of ranges #2#, #4#, and #4#, respectively, from #(0,0,0)#.

From the #z# intersection, we gain the height of the hypothetical rhomboid pyramid.From the #x# and also #y# intersections, we get**half**of every diagonal distance throughout the hypothetical base.

The volume of the entire **rhomboid pyramid** would have actually been:

#mathbf(V_"tetrahedron" = 1/3A_"base"h)#

The area of the **symmetrical rhombus base** is then *four* time the area the *each triangular portion*, which is the area enclosed by #y = 4 - 2x# and the #x# and also #y# axes.

#x# and also #y# come to be the elevation of the triangle, and we settle for that area as #A_"triangle" = 1/2xy#. Thus:

#A_"base" = 4(1/2xy) = 2xy = 2(2)(4) = 16#

Or, we can have used the formula because that the area of a rhombus ("diagonals method"), which supplies #2x# and also #2y# as the diagonals #p# and #q#.

#A_"base" = (pq)/2 = ((2x)(2y))/2 = 2xy = 16#

Finally, by construction, the **volume that the initial tetrahedron** is climate one-fourth the volume of our theoretical rhomboid pyramid:

#color(blue)(V_"tetrahedron") = 1/4<1/3Ah>#

#= 1/4*1/3<16*4>#

#= 1/4*64/3#

#= color(blue)(16/3)#

**CALCULUS III APPROACH**

An different approach to this using **triple integrals** requires integrating each measurement at a time.

#=> mathbf(int_(x_1)^(x_2) int_(y_1)^(y_2) int_(z_1)^(z_2) dzdydx)#

What we have is #x_1 = y_1 = z_1 = 0#, due to the fact that the lower bound is each coordinate plane. The is, we understand that #x,y,z >= 0#, so we room bound by those values.

Next, to acquire the upper bounds, we deal with the equation for each individual variable.

Solving for #z_2#, we gain #color(green)(z_2 = 4 - 2x - y)#.Note: ours integration facet can"t have actually #x = y = 0#, since #z = 4 - 2x# is our #xz#-plane triangle, and also #y# enables us to incorporate with respect come #y# later. This is ours **projection along the** #mathbf(y)# **axis**.

**three**dimensions, there exist

**two**intersections ~ above the #xy#-plane: as soon as #x = 0#, and also when #y = 0#. Us can encompass both of lock in one 2-variable equation once #z = 0# to get:

#color(green)(y_2 = 4 - 2x)#

Note: our integration aspect can"t have actually #x = 0#, since #y = 4# is simply a horizontal line, and also we require to combine with respect come #x# later. This is our **projection along the** #mathbf(x)# **axis**.

#2x_2 = 4 - z - y => 2x_2 = 4#

#color(green)(x_2 = 2)#

Overall, we should snapshot the #xz#-plane constructed by the #x# and also #z# intercepts, **projected outwards follow me the** #mathbf(y)# **axis**, bounded:

like so:

to create the tetrahedron:

So, our integrals job-related out like this, indigenous the inside out:

#int_(0)^(2) int_(0)^(4 - 2x) int_(0)^(4 - 2x - y) 1dzdydx#

#= int_(0)^(2) int_(0)^(4 - 2x) 4 - 2x - y dydx#

Now for the "partial" integral through respect to #y# (the station of the partial derivative v respect come #y#). So, #x# is a constant.

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#= int_(0)^(2) |<4y - 2xy - y^2/2>|_(0)^(4-2x) dx#

#= int_(0)^(2) <(4(4-2x) - 2x(4-2x) - (4-2x)^2/2) - cancel((4(0) - 2x(0) - (0)^2/2))> dx#

#= int_(0)^(2) <(16-8x) - (8x-4x^2) - (16 - 16x + 4x^2)/2> dx#

#= int_(0)^(2) <(16-8x) - (8x-4x^2) - (8 - 8x + 2x^2)> dx#

#= int_(0)^(2) 16 - 8x - 8x + 4x^2 - 8 + 8x - 2x^2 dx#

Finally, the integral through respect come #x# is easier, with only one variable to transaction with.