2

The problem is as follows: Compute the intergal $$I= \iiint_B \frac{x^4+2y^4}{x^4 +4y^4 +z^4 }\text{d}x\,\text{d}y\,\text{d}z $$ where $B$ is the unit ball defined by $B=$ {$(x,y,z)∣x^ 2 +y^ 2 +z^ 2 \leq 1$} .

The official solution is tricky: The change of variable $(x,y,z) \mapsto (z,y,x)$ transforms the integral into $$ I= \iiint_B \frac{z^4+2y^4}{x^4 +4y^4 +z^4 }\text{d}x\,\text{d}y\,\text{d}z \qquad(1)$$ hence $2I= \iiint_B 1\, \text{d}x\,\text{d}y\,\text{d}z = 4\pi/3$. Therefore, $I=2\pi/3$. My question is: what is meant by $(x,y,z)\mapsto (z,y,x)$ , isn't it ambiguous? Somebody have another solution saying that by symmetry, we have $I$ equals $(1)$ and adding gives the answer. I wanna ask if symmetric here means that the permutation of the variables preserves the domain $D$? Why if it is symmetric, then the two integrals are the same?

The next question is to compute the integral $$J=\int_0^1 \int_0^1 \int_0^1 \cos^2(\frac{\pi}{6}(x+y+z)) \text{d}x\,\text{d}y\,\text{d}z$$ The solution uses similar technique in above: substitutes $x=1-u. y=1-v, z=1-w$, then we will get $$J=\int_1^0 \int_1^0\int_1^0 \cos^2(\frac{\pi}{2}-\frac{\pi}{6}(u+v+w)) \text{d}u\,\text{d}v\,\text{d}w$$ but why it also equals to $$\int_0^1 \int_0^1\int_0^1 \cos^2(\frac{\pi}{2}-\frac{\pi}{6}(u+v+w)) \text{d}u\,\text{d}v\,\text{d}w$$ did I do something wrong?? Thank you for answering such a dumb question!

nayr
  • 131

2 Answers2

1

For your first question, the solution is not so mysterious. You are integrating over a region symmetric in $x$ and $z$. Therefore you can switch $x$ for $z$ and get the same answer. Specifically:

$$x^2+y^2+z^2=1 \implies z^2+y^2+x^2=1$$

$$\begin{align}I=\iiint_B dx\,dy\,dz \frac{x^4+2 y^4}{x^4+4 y^4+z^4} &= \iiint_B dz\,dy\,dx \frac{z^4+2 y^4}{z^4+4 y^4+x^4}\\ &= \iiint_B dx\,dy\,dz \frac{z^4+2 y^4}{z^4+4 y^4+x^4}\end{align}$$

Note that the volume element is the same after substitution. You may then add the two equal quantities as you mentioned above:

$$\begin{align}2I &= \iiint_B dx\,dy\,dz \frac{x^4+2 y^4+z^4+2 y^4}{x^4+4 y^4+z^4} \\ &= \iiint_B dx\,dy\,dz\\ &= \frac{4 \pi}{3}\\\implies I&=\frac{2 \pi}{3} \end{align}$$

Ron Gordon
  • 138,521
0

Start with the innermost integral:

$$\int_0^1{cos^2\frac{\pi}{6}(x+y+z)dx}$$

Treat all variables besides x as constants, and using the trigonometric identity, the indefinite integral is

$$\frac{x}{2}+\frac{3}{2\pi}sin(\frac{\pi}{3}(x+y+z))$$

Now plug into x the limits 0 to 1, and you get this function:

$$\frac{1}{2}-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))$$

Now you have a function that is only dependent on y and z, and for your total integral you're left with only two integrals:

$$\int_0^1\int_0^1{(\frac{1}{2}-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z)))}dy dz$$

Take the integral with respect to y of this function(y is the next outermost integral), taking all variables by y to be constant:

$$\int_0^1{(\frac{1}{2}-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z))-\frac{3}{2\pi}sin(\frac{\pi}{3}(y+z)))}dy$$

Rinse, repeat, until there's no more integrals.

rurouniwallace
  • 6,266