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?

  • $\begingroup$ Check your bounds for $x$. For $x \gt 1$, lower bound of $y$ is not a real number. Also $z$ cannot be a constant as upper bound of $z$ is not $3$ at every point in the region. Please see my answer and let me know if any questions. $\endgroup$
    – Math Lover
    Jun 16 at 6:44

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$

  • $\begingroup$ I tried to plot the equations based in your solution in a graph and all makes sense. So, I tried to setup the iterated integral with order $dzdxdy$. The z bounds is still $0 \leq z \leq \sqrt{x^2+y^2-1}$ For $0 \leq y \leq \sqrt{10}$, x is bounded between $x = \sqrt{10-y^2}$ and the y-axis. For $0 \leq y \leq 1$, x is bounded between $x = \sqrt{1-y^2}$ and the y-axis. The iterated integral is $\int^{10}_{0} \int^{\sqrt{10-y^2}}_{0} \int^{\sqrt{x^2+y^2-1}}_{0} zdzdxdy - \int^{1}_{0} \int^{\sqrt{1-y^2}}_{0} \int^{\sqrt{x^2+y^2-1}}_{0} z dzdxdy$ Is my understanding right? $\endgroup$
    – Fubuki
    Jun 17 at 0:59
  • 1
    $\begingroup$ @Aiden Yes you can either add between $0 \leq y \leq 1$ and $1 \leq y \leq \sqrt{10}$ or find for $0 \leq y \leq \sqrt{10}$ and then subtract $0 \leq y \leq 1$. But you have written in your integral $\int_0^{10}$. It should be $\int_0^{\sqrt{10}}$ $\endgroup$
    – Math Lover
    Jun 17 at 4:14
  • $\begingroup$ @Aiden sorry I missed seeing it closely earlier. Subtracting it does not work as the hyperboloid surface does not exist for $x^2 + y^2 \leq 1$. So how are you defining the upper bound of $z$? in the second integral. Can you see that $\sqrt{x^2+y^2-1}$ is not defined for $x^2 + y^2 \lt 1$? $\endgroup$
    – Math Lover
    Jun 17 at 6:13
  • $\begingroup$ Any reason you are not using the set up I gave in my answer. That is the correct way to do it. You can change the order from $dy \ dx$ to $dx \ dy$ but you cannot subtract like this. $\endgroup$
    – Math Lover
    Jun 17 at 6:15
  • 1
    $\begingroup$ In my set up when we are between $0 \leq x \leq 1$, do you see the lower bound of $y$ is $\sqrt{1-x^2}$? This makes sure that $x^2 + y^2$ is never less than $1$ and $z = \sqrt{x^2+y^2-1}$ is always defined. $\endgroup$
    – Math Lover
    Jun 17 at 6:17

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