I need to setup the triple integral in cartesian coordinates to solve the volume of $f(x, y, z) = z$ inside the cylinder $x^2 + y^2 = 10$ and outside the hyperboloid $x^2 + y^2 – z^2 = 1$ in the first octant.
What I did so far is to find the intersection between the cylinder and hyperboloid so that I can find the bounds for z. Actually, I am uncertain if finding this intersection is a right procedure.
$$ 10 – z^2 = 1 \implies z = 3 $$
My bounds for z is $0 \leq z \leq 3$ since the volume is also bounded by the first octant.
My bounds for y is $\sqrt{1-x^2} \leq y \leq \sqrt{10-x^2}$. Lastly, my bounds for x is $1 \leq x \leq \sqrt{10}$.
My iterated integral for this one will be like this
$$ \int^{\sqrt{10}}_{1} \int^{\sqrt{10-x^2}}_{\sqrt{1-x^2}} \int^{3}_{0} z dzdydx$$
Do I miss something here?
Just a picture, not an answer.
Best Answer
No that is not correct. Both $z$ bounds cannot be constant.
It is much easier in cylindrical coordinates but if you have to set it up in cartesian coordinates,
Integrating wrt $z$ first,
$0 \leq z \leq \sqrt{x^2+y^2-1}$
For $x, y$, the bounds are split.
i) For $0 \leq x \leq 1$, $y$ is bound between two circles in XY-plane ($x^2+y^2 = 1$ and $x^2+y^2 = 10$).
$\sqrt{1-x^2} \leq y \leq \sqrt{10-x^2}$
ii) For $1 \leq x \leq \sqrt{10}$, $y$ is bound between x-axis and circle $x^2+y^2 = 10$.
$0 \leq y \leq \sqrt{10-x^2}$
So the integral would be,
$\displaystyle \int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{10-x^2}} \int_0^{\sqrt{x^2+y^2-1}} z \ dz \ dy \ dx \ \ + $
$ \displaystyle \int_1^{\sqrt{10}} \int_0^{\sqrt{10-x^2}} \int_0^{\sqrt{x^2+y^2-1}} z \ dz \ dy \ dx$