<p>Can someone give me an intuitive interpretation of what a triple integral means? According to the book, it means that solid is divided into tiny boxes with volume = dx<em>dy</em>dz, and multipled by the value of f(x<em>,y</em>,z<em>), with (x</em>,y<em>,z</em>) lying INSIDE the box…summing up the above product does NOT equal to volume, right? but that’s the definition of the triple integral. So what’s an intuitive interpretation of a triple integral? My TA is way too incoherent for me to understand so I’m posting the question here…</p>
<p>Why doesn’t it equal the volume? What you just said IS the volume… By the way, this isn’t a homework help site.</p>
<p>A triple integral as an intergral with 3 dependent variables. Volume has 3 variables, so what’s the problem?</p>
<p>Nevermind, I got the answer. I just posted it on this forum so a few engineering students could help me out.</p>
<p>By the way, lil_killer and Japher, a triple integral does not automatically equal volume. It’s the double integral that will always give you the volume. The triple integral of f(x,y,z)dV actually gives you the hypervolume of a 4 dimensional object. It only gives you the volume when f(x,y,z) = 1</p>
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<p>Nope. For a function of 2 variables, z = f(x,y), the double integral gives you the volume, because when one takes the sum of f(x,y)dA, one is summing up the volume of all the tiny columns one has divided the solid into. This is so because f(x,y), or z, represents the height of the column above the small rectangle with area dA = dydx, and multiplying this area by the height z, gives you the volume of the column, and therefore, if you sum up all those individual volumes, you get the double integral of f(x,y)dA. This is NOT so for the triple integral. Even my book says that the triple integral of f(x,y,z) should not be automatically interpreted as a volume, because f actually lies in a FOUR dimensional space, not a 3D space, since its a function of 3 variables.</p>
<p>If your function is equal to 1, then you’d get the volume. </p>
<p>I’m using f as the integral symbol.</p>
<p>ff dA = ff dxdy= area
fff dV = fff dxdydz = volume</p>
<p>^ yeah, what steevee said, I really have no idea what this guy is getting at</p>
<p>Say for example the density of an object changes depending on where in the object you are, you would use a triple integral to find the mass of that object. Mass = Integral of density as a function of space times your differential spacial coordinates. Mass=fff(rho(x,y,z)<em>dx</em>dy*dz)</p>
<p>Another example, say you want to know the volume-average temperature of an object where there is some internal temperature gradient as a function of x,y, and z coordinates then, you could use a triple integral: Average Temp=(1/Volume)fff(T(x,y,z)<em>dx</em>dy*dz)</p>