Mass of body between curve and plane

Question

Let $R$ be the region of the parabloid $z=4-x^2-y^2$ where $z \geq 3$ and let $D$ be the body bounded by $R$ and the plane $z=3$.

Setting $z=z$ I obtain $x^2+y^2=1$ which is the $xy$-projection I assume. The density $p$ is constant $ = 1$

Is this the correct setup?: $\int_0^{2\pi} \int_0^1\int_3^4 1 dzdydx$

Answer 1

Hint:

You can conclude your work using cylindrical coordinates: the region is defined by the limits: $$ 3\le z\le 4-r^2 \qquad 0\le r \le 1 \qquad 0\le \theta\le 2\pi $$

so the volume is given by: $$ V=\int_0^{2\pi}\int_0^1 \int_3^{4-r^2} r\;dzdrd\theta $$

Answer 2

No, your limits are off. $$\int_{-1}^{1} \int_{- \sqrt {(1-X^2)} }^{ \sqrt {(1-X^2)} } \int_3^{4-x^2-y^2} dzdydx$$

That is what you want.

Or you can use polar which is a different story.