I need to find the volume of the solid bounded by the surfaces $ + − 1 = 0$ , $ = 1$ and $^2 + ^2 = 4$ in $\mathbb{R}^3$.
However, when I use the triple integral: $$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{0}\int_{1}^{1-x}dz~dy~dx$$, I get the answer $0$. Therefore, I have to find the volume using another integral: $$2\int_{-2}^{0}\int_{-\sqrt{4-x^2}}^{0}\int_{1}^{1-x}dz~dy~dx$$ which gives me the answer $\frac{16}{3}$. (I am not sure this is correct.)
Am I taking the right limits for the integrals?
On
As I wrote in comment it's better to separate figure with respect to $z=1-x$ and $z=1$ interdependence, thanks for the great drawing from TeM, we have $$\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-y^2}}\int\limits_{1-x}^{1}dzdxdy+\int\limits_{-2}^{2}\int\limits_{-\sqrt{4-y^2}}^{0}\int\limits_{1}^{1-x}dzdxdy = \frac{16}{3}+\frac{16}{3}= \frac{32}{3}$$
Drawing the surfaces of Cartesian equation:
$$ z = 1, \quad \quad \quad \quad z = 1 - x, \quad \quad \quad \quad x^2 + y^2 = 4 $$
$\quad\quad\quad\quad\quad$
it follows that the volume of the enclosed region of space is equal to:
$$ V = 2\cdot\iiint\limits_{\{x<0,\,1<z<1-x,\,x^2+y^2<4\}} 1\,\text{d}x\,\text{d}y\,\text{d}z = 2\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \text{d}\theta \int_0^2 \rho\,\text{d}\rho \int_1^{1-\rho\cos\theta} 1\,\text{d}z = \boxed{\frac{32}{3}} $$