Working out the limits of a triple integral - a volume bounded by a plane in the first octant

Question

I'm attempting to do the following exercise:

And am now fairly aware now that the hardest part about solving these types of problems is understanding the bounds you are integrating in.

Firstly, I try and draw a horrendous sketch of what I believe we're integrating in. I note:

• Intercepts: $x = h/a$, $z=h$, $y=h/b$, found by setting two variables equal to $0$ and solving for the other one.

• With $x=0$, $z = -by + h$.

• With $y=0$, $z = -ax + h$.
• With $z=0$, $y = -ax/b + h/b$

Keeping this in mind, here is the sketch I drew out what I think this is:

Now, where the confusing part comes in (assuming I'm making sense so far). I need to note the order of integration I'd find would be most straightforward. My plan of attack is to make a line from the ends of $x$ within the object at $z=0$, which would involve integrating $x$ first, then moving it along $y$, so that it feels the trapezoid-looking figure at $z=0$. Then, I'll integrate in $z$, and have the entire volume filled. So, I'll try and establish some limits.

For $x$, our first line goes from $by = -ax +h$, which arranges for $x$ to $x=-\frac{b}{a}y+h/a$ to the minimum value of $x$, which is the intercept $x=h/a$.

For $y$, our line is scaled in $y$ from the $xz$ plane to the $yz$ plane to fill the area of the base of this figure.

Finally, for $z$ our volume is made from scaling the area of the trapezoid from 0 to $h$.

Thus, our limits are:

$$-(\frac{b}{a}y+h/a) \le x \le h/a$$

and for $y$..

according to my thinking, it integrates from $z = -ax+h$ to $z = -by + h$ which means one of my limits cannot be expressed in terms of $y$ and $z$ which means I can't get a integrate the final $z$ integral properly since I'll have an $x$ term in it.

Where am I going wrong? Is the sketch wrong? Are my thought processes wrong? If so, which? I have a feeling my sketch is wrong, as this object doesn't exactly strike me as a plane. Also, the bottom figure is not necessarily a trapezoid, as $b \ne a$ necessarily.

Answer 1

If you just mark where the intercepts are and then connect them to form the plane itself, then that's usually sufficient. For instance, based on what you have here, I can say that:

$z$ is bounded below by $0$ and above by $h - ax - by$.

Then $y$ is bounded at the "back" by $0$ and at the "front" by $-ax/b + h/b$.

Finally $x$ is bounded to the left by $0$ and to the right by the intercept, i.e. $h/a$.

So I believe your integral should look like $$\int_0^{h/a} \int_0^{-ax/b + h/b} \int_0^{h - ax - by} z \, \mathrm dz \, \mathrm dy \, \mathrm dx$$.

Answer 2

The volume $V$ is described by $0\le ax+by+z\le h$

Let's isolate $z$.

$a,b,x,y\ge 0$ means we keep the lower bound, so $0\le ax+by\le h-z$

• $z\ge 0$ and $h-z\ge 0$ so $z\in[0,h]$.

We can now isolate $y$.

$a,x\ge 0$ means we still keep that lower bound, so $0\le ax\le h-z-by$

• $y\ge 0$ and $h-z-by\ge 0$ so $y\in[0,\frac{h-z}b]$

Finally we finish with $x$

• $x\in[0,\frac{h-z-by}a]$

Overall the integral becomes

$I=\displaystyle\iiint_V z\mathop{dx}\mathop{dy}=\mathop{dz}=\int_0^h\left(\int_0^{\frac{h-z}b}\left(\int_0^{\frac{h-z-by}a}\mathop{dx}\right)\mathop{dy}\right)z\mathop{dz}=\dfrac{h^4}{24ab}$