Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am struggling a lot with triple integrals. I can evaluate them, but I find it extremely difficult to write the triple integral. I cannot visualize them.

An example question: $\iiint_E6xy\;dV$, where $E$ lies under the plane $z = 1 + x + y$ and above the region in the xy-plane bounded by the curves $y=\sqrt{x}$, $y=0$, $x=1$.

Since the solid lies under the plane, the upper limit for $z$ would be $1 + x + y$. Since the lower bounded region is in the xy plane, the lower bound would be 0? Thus, $\iiint_0^{1+x+y}6xy\;dz...$

For the other integrals x and y, I would simply sketch the xy region and integrate the same as double integrals. Which I get: $\int_0^1\int_{\sqrt{x}}^1\int_0^{1+x+y}6xy\;dz\;dy\;dx$

Are these limits of integration correct? Is the approach correct? The professor taught by visuals, which I find extremely difficult to do in an xyz plane.

share|cite|improve this question
up vote 1 down vote accepted

Try starting by just sketching the projection into the $xy$-plane, which seems to be what you're thinking. However, the bounds should be $0 \leq y \leq \sqrt x$ , $0 \leq x\leq 1 $. One way to see this is that this region in the $xy$-plane is bounded by two curves that are each described by $y$ as a function of $x$. Now you can look at this region like you would in calc. 2: draw a vertical rectangle from the lower curve ($y=0$) to the upper curve ($y=\sqrt x$ ).

With this in mind, you should get the following:


share|cite|improve this answer
Thank you. I drew my X/Y plot a bit wrong, which is why I got my dy mixed. This is a bit of a big, ugly problem still (a lot of places to mess up your calculations), so I still get some wrong calculations, but it's reassuring to know I at least have my limits of integration correct, which is definitely the most important part. – user1405177 Oct 28 '12 at 17:01

Denote points in ${\mathbb R}^{p+q}$ by $({\bf x},{\bf y})$, $\ {\bf x}\in{\mathbb R}^p$, $\ {\bf y}\in {\mathbb R}^q$, and denote the projection onto the ${\bf x}$-plane by $\pi$: $$\pi:\quad{\mathbb R}^{p+q}\to{\mathbb R}^p,\qquad({\bf x},{\bf y})\mapsto{\bf x}\ .$$ Assume that we are given a beautiful body $B\subset {\mathbb R}^{p+q}$ and a nice function $f:\ B\to{\mathbb R}$.

The set $B$ has a projection $$B':=\pi(B)=\{{\bf x}\,|\, \exists {\bf y}\in{\mathbb R}^q:\ ({\bf x},{\bf y})\in B\}$$ onto the ${\bf x}$-plane, and for each ${\bf x}\in B'$ there is the set $B_{\bf x}$ of points ${\bf y}\in{\mathbb R}^q$ such that $({\bf x},{\bf y})$ is a "body point" projecting onto the given ${\bf x}$: $$B_{\bf x}=\{{\bf y}\in{\mathbb R}^q\,|\, ({\bf x},{\bf y})\in B\}\ .$$ The "theorem of Fubini" then says that $$\int\nolimits_B f({\bf x},{\bf y})\ {\rm d}({\bf x},{\bf y})= \int\nolimits_{B'}\left(\int_{B_{\bf x}}f({\bf x},{\bf y})\ {\rm d}({\bf y})\right)\ {\rm d}({\bf x})\ .\qquad(*)$$ Now this is a general principle. In your case $$f(x,y,z)=6xy\ ,$$ the projection $\pi$ is $(x,y,z)\mapsto (x,y)$, and $$B'=\{(x,y)\,| 0\leq x\leq 1,\ 0\leq y\leq\sqrt{x}\} ,\qquad B_{(x,y)}=[0,1+x+y]\ .$$ The "general principle" $(*)$ should not be for you a mysterious formula: If you approximate the integrals occurring therein by Riemann sums in terms of tiny boxes of $p$-dimensional "width" and $q$-dimensional "height" it is intuitively obvious.

share|cite|improve this answer
$I$ like your answer but it seems to be a bit much for the $original$ poster's question, whom I'm assuming is taking calc. 3/multivariable calculus. – BobaFret Oct 28 '12 at 19:06

Start by drawing the region in the $xy-$plane. This will look like a right-angled triangle, but with one side "bent" outward. I.e. the region $0\leq y \leq \sqrt{x}$ intersected with $0\leq x \leq 1.$

Now, imagine a "tower" with a base like this shape, extending upwards. This tower is then cut by a slanted plane, the $1+x+y = z$ plane.

For example, the height of the tower in the corner $(x,y)=(0,0)$ will have height $1,$ the corner $(1,1)$ height $3$ and $(1,0)$ height 2.

This feels like a sufficiently good picture to me. Now, the integrand $6xy$ represents some density of the building material. The lower left corner has density 0, so imagine a very light material, and it increases gradually until the density is 6, in the top right corner. From this, we can see that the integral itself must be a number between 0, and approximately $6\cdot (3 \cdot \frac12) = 9$, which is the maximal density times the approximate base area (I underestimate the area a bit, but the density is grossly over-estimated.).

Hope it helps!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.