- #1

mathmari

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MHB

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We have the set $M=\{(x,y,z)\in \mathbb{R}^3 : x^2+y^2-z^2\leq 1, \ 0 \leq z\leq 3\}$. Draw $M$ and calculate the volume of $M$. I have done the following :

\begin{equation*}\int_M\, dV=\int\int\int\, dx\, dy\, dy\end{equation*} Which are the boundaries of the integrals? Do we have to use spherical coordinates?

Or do we set $x=r\cos\theta$ and $y=r\sin\theta$ and $z$ remains $z$ with $0\leq z\leq 3$ ?

We have that $x^2+y^2-z^2\leq \Rightarrow r^2-z^2\leq 1 \Rightarrow r^2\leq 1+z^2 \Rightarrow -\sqrt{1+z^2}\leq r\leq \sqrt{1+z^2}$ and since $r$ is the radius and so it must be positive we get $0 \leq r\leq \sqrt{1+z^2}$, right? But what about $\theta$ ?

Or is there an other way to calculate that?

:unsure: