0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $$V = \int_0^{2\pi}\int_0^1 \int_0^{4-x^2-y^2} r\, dz\, dr\, d\theta$$

$\endgroup$
0
$\begingroup$

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 $y=0$.

enter image description here

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.