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Evaluate the triple integral $$\iiint_E y ~dV$$ where $E$ is bounded by the planes $x=0$, $y=0$, $z=0$, and $2x + 2y + z = 4$.

How we can find other boundaries for $x$, $y$, $z$? So far, we have $0 \le x, 0 \le y, 0 \le z$.

Is there a way to find those boundaries without sketching the graph ?

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I don't know how to explain it without a blackboard. Here is a (poor) try.

One has to keep constantly in mind is that $x$, $y$, and $z$ must always be $\ge 0$.

One should also try to visualize the region. It is pyramidal, with triangular faces.

The $z$ is easiest. Since our plane has equation $2x+2y+z=4$, the variable $z$ goes from $0$ to $4-2x-2y$, as long as $4-2x-2y$ is positive. But that will be taken care of later.

Now that $z$ has been taken care of, we need to integrate over the base (in the $x$-$y$ plane) of the solid. So our problem is comfortably $2$-dimensional. Let's deal with $y$. As usual, we need $y \ge 0$. At the top, we need to make sure $4-2x-2y\ge 0$. So we need $2y \le 4-2x$, or equivalently $y\le 2-x$.

Finally, we need $2-x \ge 0$. So $x$ goes from $0$ to 2$.

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  • $\begingroup$ +1 You did it great without a blackboard. This is what I usually need here. $\endgroup$ – mrs Nov 28 '12 at 7:14
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I am sure there are ways of finding bounds without sketching the graph, but they involve a lot of inequalities. Just trying to do so for a 2D region with a single integral takes quite a bit of computing. (I covered the basic idea for another user through that in this post (didn't compute, but just gave concept): Area Between Curves)

I highly recommend NOT trying to find bounds without graphing. The graph in question is a simple tetrahedron--it's not hard at all. Difficult bounds to sketch aren't (typically) asked of students unless they have graphing calculators. In the "real world" you have access to computer algebra systems, so you can use that to find the bounds.

tl;dr: Don't. It opens up too much opportunity for error.

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