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?
 A: 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 $$\int_Tdxdydz=\int_{S(T)}\Bigg|\frac{\partial(x,y,z)}{\partial(u,v,w)}\Bigg|dudvdw=\frac{\text{Volume of } S(T)}{2}=\frac{3\pi}{2}$$
A: 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:
$$V=\int_0^3 S(z)dz=\frac{3\pi}{2}$$
A: 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 :
$$0\leq z\leq3 \\ -\frac{z}{2}-\frac{1}{\sqrt{2}} \leq x \leq -\frac{z}{2}+\frac{1}{\sqrt{2}} \\-\frac{z}{2}-\sqrt{\frac{1}{2}-\left(x+\frac{z}{2} \right)^2}\leq y\leq -\frac{z}{2}+\sqrt{\frac{1}{2}-\left(x+\frac{z}{2} \right)^2} $$
