find the volume of a solid limited by a paraboloid and a cylinder

Question

Find the volume of a solid inside the paraboloid $z=4-x^{2}-y^{2}$ and the cylinder $x^2+y^2=1$ and above $xy$-plane

How to start? How to find the limits of the integration?

Answer 1

The region projected onto the $x-y$ plane gives a circle of radius $1$, and $z$ is integrated from $0$ to $4-x^2-y^2$. Therefore, using cylindrical coordinates, the integral becomes $$V = \int_0^{2\pi}\int_0^1 \int_0^{4-x^2-y^2} r\, dz\, dr\, d\theta$$

Answer 2

Another approach: the volume (of revolution) consists of two part - the cylinder for $0\le z\le 3$ and the top for $3\le z\le 4$. The sketch is for $y=0$.

The first volume (the cylinder) is simply $3\pi$. The second one is (by disc method) $$ \int_3^4\pi x^2\,dz $$ where $z=4-x^2$ $\Leftrightarrow$ $x^2=4-z$.

Answer 3

Draw a sketch at first. Rhere is circular symmetry.The volume of a paraboloid is one half that of enclosing cylinder. By by disc method

$$\int_0^a 2 \pi r dr= \pi a^2/2 $$

For a cylinder the volume is $$ \pi a^2 h$$

So total volume is

$$ \pi \cdot 1^2 \cdot \frac12 + \pi \cdot 1^2\cdot 3 $$