# Finding the volume of $\{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$ by integration

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?

• oblique cylinder with its axis $\frac{x}{1} = \frac{y}{1} = - \frac{z}{2}$ and radius of $\frac{1}{\sqrt2}$. – 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 $$\int_Tdxdydz=\int_{S(T)}\Bigg|\frac{\partial(x,y,z)}{\partial(u,v,w)}\Bigg|dudvdw=\frac{\text{Volume of } S(T)}{2}=\frac{3\pi}{2}$$

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

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:

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

• Why do you think your result is different than the one of @Matthew Holder? – qcc101 Oct 29 at 16:02
• 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. – qcc101 Oct 29 at 16:07
• I had made an error that I have corrected now. Sorry. I find indeed the same result as him. – 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 :

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

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