# How to visualize triple integrals?

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.

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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:

$$\int_0^1\int_{0}^{\sqrt{x}}\int_0^{1+x+y}6xydzdydx$$

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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.

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$I$ 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!

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