What's the easier aproach to calculate the volume of the following region:
$$\{(x,y,z) \in \mathbb{R}^3: 0\leq y \wedge x+y\leq 1 \wedge x^2 + y^2 + z^2 \leq 1 \}$$
The way I did this will sound really dumb, and I got a really complicated integral so solve, which is not the idea of the exercise.
I restricted my problem to $z \geq 0$, there is symmetry. Then I wanted to study the region where $0\leq z \leq \frac{\sqrt{2}}{2}$ in which the plane defined by $x+y \leq 1$ intersects the level curves $x^2+y^2= 1 - z^2_0$ in two points.
From here i was able to define the area define the "area of the region" in $z_0$. Which if I didn't get it wrong, it is given for a fixed $z$: $\arctan(-1+\frac{1}{z^2}-\frac{\sqrt{1-2z^2}}{z^2}) + \frac{\sqrt{4-8z^2}}{4}$
Integrating this doesn't sound like the intention of the exercise. So any solution or hint would be helpful
As $x+y \leq 1, y \geq 0, x^2+y^2+z^2 \leq 1$,
Bounds of $z$ is defined by the sphere so $ - \sqrt{1-x^2-y^2} \leq z \leq - \sqrt{1-x^2-y^2}$.
Lower bounds of $x$ is sphere and upper bound is plane $x + y \leq 1$.
So, $ - \sqrt{1-y^2} \leq x \leq 1-y$
Finally, bounds of $y$ is $0 \leq y \leq 1$.
So integral to find volume is
$\displaystyle V = \int_0^1 \int_{- \sqrt{1-y^2}}^{1-y} 2 \sqrt{1-x^2-y^2} \ dx \ dy$
Another approach is to rotate coordinate system clockwise by $\frac{\pi}{4}$ around z-axis and use cylindrical coordinates.
$x = r \cos (\theta - \frac{\pi}{4}), y = r \sin (\theta - \frac{\pi}{4}), r^2 + z^2 \leq 1$
$x + y \leq 1 \implies \sqrt2 \ r \sin\theta \leq 1$
So volume of the region is given by,
$\displaystyle V = \int_{3\pi/4}^{5\pi/4} \int_0^1 2 r \sqrt{1-r^2} \ dr \ d\theta + \int_{\pi/4}^{3\pi/4} \int_0^{\csc\theta / \sqrt2} 2 r \sqrt{1-r^2} \ dr \ d\theta$
$ = \displaystyle \frac{5 \pi}{6 \sqrt2}$