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"A parallelepiped is described by the vectors $(2,2,3),(2,4,3)$ and $(0,1,5)$

given that the density $= 2x+2y$, find the mass of the parallelepiped."

I can find the volume just fine, but setting up the integral to find the mass is giving me a lot of trouble.

this is my best guess so far...

$$20 \int_0^2 \int_0^4 \int_0^5 (2x+2y) dzdxdy$$

The vectors do stem from the origin

Kelvin Lois
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Zac Zaccardi
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    mass = volume times density – Vasili Mar 29 '18 at 01:10
  • Why don’t you show your efforts at setting up this integral so that someone can help you with the specific things that are giving your trouble? – amd Mar 29 '18 at 01:16
  • if the density was a whole number, that would be pertinent. However, the density is a function of x and y, therefore I need to integrate that function over unknown bounds. what would those bounds be? – Zac Zaccardi Mar 29 '18 at 01:17
  • You are probably integrating $\int \int \int 1dz dy dx$ or something similar, because as Vasya pointed out density*volume = mass you can multiply the 1 by the density function. – futurebird Mar 29 '18 at 01:17
  • I think it might be easier to translate both functions so that the parallelepiped can be described with a single vector from the origin. – futurebird Mar 29 '18 at 01:22
  • Transform to a coordinate system in which the bounds are all $[0,1]$. – amd Mar 29 '18 at 01:22
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    You’re not integrating over the given paralellepiped but instead over a box with sides parallel to the coordinate axes. Moreover, the two regions don’t even have the same volume. – amd Mar 29 '18 at 01:27
  • well, I don't know how to go about doing that. You can imagine why im having such trouble with this problem then. – Zac Zaccardi Mar 29 '18 at 01:32
  • Never mind about translating it. – futurebird Mar 29 '18 at 01:41
  • I think what you have is correct, Zac... but I'm very rusty. But, what you would be doing is summing the value of the density function across the volume of the solid. So it makes conceptual sense.

    Why is 20 there?

     – futurebird Mar 29 '18 at 01:59
  • sorry, the 20 is the volume calculated by taking the det of the three vectors. I assumed by the above explanations that the mass could be calculated by multiplying the calculated volume by the mass given by the integral, but now i realize that it wouldnt give me the correct answer – Zac Zaccardi Mar 29 '18 at 02:01

1 Answers1

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You’re starting out with the right idea: you need to compute the triple integral $$\iiint 2x+2y\,dV$$ over the paralellepiped, but since the edges aren’t parallel to the coordinate axes, finding the right bounds with $dV=dx\,dy\,dz$ is a bit of a nuisance. So, perform a change of variables to a coordinate system in which the edges are parallel to the coordinate axes. One possibility is the transformation $$(x,y,z) = (2,2,3)u+(2,4,3)v+(0,1,5)w = (2u+2v,2u+4v+w,3u+3v+5w)$$ so that each variable falls in the range $[0,1]$. The integral then becomes $$\int_0^1\int_0^1\int_0^1 2(2u+2v)+2(2u+4v+w)\,\left|{\partial(x,y,z)\over\partial(u,v,w)}\right|\,du\,dv\,dw.$$ Since the coordinate transformation is linear, its Jacobian is just the determinant of its matrix, which in turn is equal to the volume $V$ of the paralellepiped, so the integral becomes $$2V \int_0^1\int_0^1\int_0^1 4u+6v+w\,du\,dv\,dw$$ which I’m sure you can evaluate.

amd
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    Alright, that makes sense to me. Though the answer I get is none of the possible answers given to me in my homework. I trust that to be an error by my professor( which is very common...). Thank you! – Zac Zaccardi Mar 29 '18 at 03:02