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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?

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  • $\begingroup$ 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, 2020 at 15:03

3 Answers 3

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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 $$\int_Tdxdydz=\int_{S(T)}\Bigg|\frac{\partial(x,y,z)}{\partial(u,v,w)}\Bigg|dudvdw=\frac{\text{Volume of } S(T)}{2}=\frac{3\pi}{2}$$

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  • $\begingroup$ I get $-\frac12$ for the Jacobian. $\endgroup$
    – saulspatz
    Oct 29, 2020 at 15:14
  • $\begingroup$ You're correct, thank you $\endgroup$
    – Matthew H.
    Oct 29, 2020 at 15:16
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Lt us integrate by slicing. The slice at height $z$ is defined by:

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

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:

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

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  • $\begingroup$ Why do you think your result is different than the one of @Matthew Holder? $\endgroup$
    – qcc101
    Oct 29, 2020 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, 2020 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, 2020 at 16:14
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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 :

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

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  • $\begingroup$ 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, 2020 at 16:12

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