I have been trying to solve this integral for some time now without success:

$$ T = \{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$$ $$ I = \int_{T} ~dx~dy~dz $$

I have tried simplifying the domain, by writing: $$ (x+y+z)^2 + (x-y)^2 \leq 1 \to $$ $$ 2x^2 + 2y^2 + z^2 + 2xz + 2yz \leq 1 \to $$ $$ x^2 + y^2 + \frac{z^2}{2} + xz + yz \leq \frac{1}{2} $$ Now, I tried doing this change of variable: $ x = \rho \cos\theta$, $y = \rho \sin\theta$ and $z = \sqrt 2 w$ With $\rho \in (0,\infty)$, $\theta \in [0,2\pi]$. They are basically cylindrical coordinates. I do some calculation and I have the following: $$ \rho^2 + w^2 + (\rho \cos\theta + \rho \sin \theta) \cdot \sqrt{2}w \leq \frac{1}{2}$$ But I don't see how this is helping. I have also tried spherical coordinates substitution without success.

Can I have some hints?

  • $\begingroup$ oblique cylinder with its axis $\frac{x}{1} = \frac{y}{1} = - \frac{z}{2}$ and radius of $\frac{1}{\sqrt2}$. $\endgroup$ – Math Lover Oct 29 at 15:03

Define the transformation $S$ by $$S(x,y,z)=(x+y+z,x-y,z)=(u,v,w)$$ Then the image of your set $T$ under $S$ is the truncated cylinder $$S(T)=\{u^2+v^2 \leq1\}\cap\{0\leq w\leq3\}\cap \mathbb{R}^3$$ It's easy to see how $$\frac{\partial(x,y,z)}{\partial(u,v,w)}=-\frac{1}{2}$$ Therefore $$\int_Tdxdydz=\int_{S(T)}\Bigg|\frac{\partial(x,y,z)}{\partial(u,v,w)}\Bigg|dudvdw=\frac{\text{Volume of } S(T)}{2}=\frac{3\pi}{2}$$

| cite | improve this answer | |
  • $\begingroup$ I get $-\frac12$ for the Jacobian. $\endgroup$ – saulspatz Oct 29 at 15:14
  • $\begingroup$ You're correct, thank you $\endgroup$ – Matthew Pilling Oct 29 at 15:16

Lt us integrate by slicing. The slice at height $z$ is defined by:


This inequation defines a disk with radius $\sqrt{\frac12}$

Therefore the area of the slice at height $z$ is constant, equal to:

$$S(z)=\pi \frac{1}{2}$$

giving a volume:

$$V=\int_0^3 S(z)dz=\frac{3\pi}{2}$$

| cite | improve this answer | |
  • $\begingroup$ Why do you think your result is different than the one of @Matthew Holder? $\endgroup$ – qcc101 Oct 29 at 16:02
  • $\begingroup$ I don't get this point, in his solution he says: z = w and z in [0,3] by hypothesis, so why is it in [0,1]? I get your solution but if I had not looked at it I would have thought he was right. $\endgroup$ – qcc101 Oct 29 at 16:07
  • $\begingroup$ I had made an error that I have corrected now. Sorry. I find indeed the same result as him. $\endgroup$ – Jean Marie Oct 29 at 16:14

Starting from your last inequation on x,y,z, you can factorize to get a cannonical expression :

$$\left(x+\frac{z}{2}\right)^2+\left(y+\frac{z}{2}\right)^2\leq\frac{1}{2}, 0\leq z\leq3$$

So you have :

$$0\leq z\leq3 \\ -\frac{z}{2}-\frac{1}{\sqrt{2}} \leq x \leq -\frac{z}{2}+\frac{1}{\sqrt{2}} \\-\frac{z}{2}-\sqrt{\frac{1}{2}-\left(x+\frac{z}{2} \right)^2}\leq y\leq -\frac{z}{2}+\sqrt{\frac{1}{2}-\left(x+\frac{z}{2} \right)^2} $$

| cite | improve this answer | |
  • $\begingroup$ I have this with my factorization : $$ x^2+y^2+z(x+y)+\frac{z^2}{2}\leq \frac{1}{2} \\ \left( \left(x+\frac{z}{2}\right)^2-\frac{z^2}{4}\right) +\left( \left(y+\frac{z}{2}\right)^2-\frac{z^2}{4}\right)+\frac{z^2}{2}\leq \frac{1}{2} \\ \left(x+\frac{z}{2}\right)^2+\left(y+\frac{z}{2}\right)^2 \leq \frac{1}{2} $$ $\endgroup$ – NHL Oct 29 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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