Setting up limits for triple integral in cylindrical coordinates

Question

I want to calculate the volume of this integral by using cylindrical coordinates $$T:Z \le2-x^2-y^2, \ Z^2 \ge x^2+y^2$$ First i want to ask here i don't see the equation for the circle i mean i don't have for example $x^2+y^2=1 \ or \ x^2+y^2=2x.$ etc. So this is my first confusion and the second one i am not sure if i am getting the limits right for the limits of integration for Z its obvious i just substitute the equations with the parameters $x=r\cos\theta, \ y=r\sin \theta$. To find the limits for r is it right to intersect the two functions ? I am doing so and i get $r=2-cos\theta-sin\theta$, so i want to ask if this mean that i must search for where $(\sin\color{blue}\theta,cos\color{blue}\theta)$ are negative on the unit circle to get the bounds for $\color{blue}\theta$ ? I found the following limits : $$r \le Z \le2-r^2 \\ 0 \le r \le2-\cos\theta-sin\theta \\ \frac{3\pi}{2} \le\theta \le\pi$$

Answer 1

The intersection of the two surfaces is found by solving $$ x^2+y^2=2-x^2-y^2$$ which is the circle $$ x^2+y^2=1$$

Thus the region of integration is the disk $$ 0\le r \le 1, 0\le \theta \le 2\pi $$ And the integrand is $$2-2r^2$$

Answer 2

HINT

Note that you have $$ x^2+y^2 \le z \le 2 - x^2-y^2, $$ and the projection of this into the $xy$-plane occurs where the boundaries intersect, i.e. when $$ x^2+y^2 = 2 - x^2 - y^2 \iff x^2+y^2=1, $$ so we will have to integrate this over the disk $D$ of radius $1$.

Can you now set up the integration?