Finding upper and lower bounds for triple integral $ \iiint_J x \,dV. $

Question

Let $D$ be the subset of $R^3$ given by $ x\ge 1$ , $y\ge 4$ and $xy\le z\le 12$.

Need to compute $ \iiint_J x \,dV$.

First I need to find the bounds where $ \int_{l_1}^{u_1} \int_{l_2}^{u_2} \int_{l_3}^{u_3} x\,dz\,dy\,dx. $ So the order is $dzdydx$

I'm struggling to find the bounds. So far I have some thoughts:

since $x\ge 1$ and $y\ge 4$, $xy\ge 4$, so $4\le z\le 12 $.

And $x\ge 1$ and since it is less than $12\ge z$, $1\le x\le 12$.

And $y\ge 4$ and since it is less than $12\ge z$, $4\le y\le 12$.

I'm not sure if this is right and how I should approach solving for the bounds. I'm trying to do it without drawing a graph, but if making one helps then I'm open to it.

Answer 1

You must start with the interior-most integration, and ask, what are the bounds placed on $z$ by the domain: $\{(x,y,z)\in\Bbb R^3: 1\leq x~, 4\leq y~, xy\leq z\leq 12\}$?

  We simply bound $z$ by $xy\leq z\leq 12$.

The logical next step is: to move outwards, having "integrated out" $z$.

Thus for the middle integration, we need $y$ to satisfy $4\leq y$ and $xy\leq 12$.

  Thus we bound $y$ by $4\leq y\leq 12/x$ .

Likewise, the outermost integration must be bounded by $1\leq x$ and $4\leq 12/x$.

  That is bound $x$ by $1\leq x\leq 3$.

$$\int_1^3\int_4^{12/x}\int_{xy}^{12} x\,\mathrm d z\,\mathrm d y\,\mathrm dx $$

Answer 2

You should sketch the region.

Intersection of plane $z = 12 ~ $ with surface $ ~ z = xy ~ $ is curve $ ~ xy = 12$.

At $x = 1, y = 12$ and at $y = 4, x = 3$. Now note that for any $x$ and $y$ in $1 \leq x \leq 3, 4 \leq y \leq 12$, $xy \leq z \leq 12$. So it is easier to set up the innermost integral with respect to $dz$ first.

Next integral can be either with respect to $dx$ or $dy$ with lower bound as $x = 1$ or $y = 4$ and upper bound being on the curve $xy = 12$.

That leads to the integral,

$ \displaystyle \int_1^3 \int_4^{12/x} \int_{xy}^{12} x ~ dz ~ dy ~ dx$

OR

$ \displaystyle \int_4^{12} \int_1^{12/y} \int_{xy}^{12} x ~ dz ~ dx ~ dy$

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Also, here is one of the orders of integral with $dz$ as the outermost integral.

$ \displaystyle \int_4^{12} \int_4^{z} \int_{1}^{z/y} x ~ dz ~ dy ~ dx$

You can work out other orders for learning purpose.