Express the volume of region $D$ upper bounded by the sphere $x^2+y^2+z^2=2$ and the paraboloid $z=x^2+y^2$.
a- Cartesian Coordinates
b- Cylindrical Surface
c- Spherical coordinates
If you help me to write the integral for a, b and c ; I can solve this integral but I made something wrong. Results are different but I don't know which result is correct.
Now, from drawing the curves, it can be seen that the region is lower bounded by the parabola, and upper bounded by the sphere. The plane of intersection is $z = 1$, which can be obtained from solving the quadratic $z^2 + z - 2 = 0$
Now, for the region below the intersection plane
$$V_1 = \int\int\int dV$$
With $0 \leq z \leq 1$, $-\sqrt{z} \leq y \leq \sqrt{z}$, $-\sqrt{z-y^2} \leq x \leq \sqrt{z-y^2}$
Hence
$$V_1 = \int_0^1 \left(\int_{-\sqrt {z}} ^ {\sqrt z}\left(\int_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}}dx\right)dy\right)dz$$
The portion above the plane of intersection will be bounded by the sphere, so take the limits accordingly