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This is the graphical setup for my problem Math3d. I given a sphere in cylindrical coordinates as the following \begin{equation}r^2+z^2=20 \end{equation} Which is the equation of a sphere of radius $\rho=2\sqrt{5}$. The problem has to be solved in the following manner (which is cylindrical systems): \begin{equation}\iiint_G f(r,\theta,z)dzdrd\theta \end{equation} The equation for the elliptical paraboloid is the following in cylindrical coordinates system. \begin{equation}z=r^2\end{equation} The problem that needs to be solved at hand is to find the volume inside the surface: \begin{equation} r^2+z^2=20 \end{equation} The volume is constrained to no being above the following surface as well: \begin{equation}z=r^2\end{equation}The question I have involves in how I set up my integral which was the following: \begin{equation} \int_0^{2\pi} \int_0^2 \int_0^{r^2} r^3dz dr d\theta \end{equation} I had hoped whether it was correct, however, there were doubts, but alas I shall post my own answer.

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  • $\begingroup$ What’s the first equation in each pair for? $\endgroup$ – amd May 22 at 17:22
  • $\begingroup$ @amd You do know that Cartesian is $(x,y,z)$, and Cylindrical is $(r,\theta,z)$? $\endgroup$ – EnlightenedFunky May 22 at 17:29
  • $\begingroup$ And where in your question did you say that you’re giving the equations in in both cylindrical and rectangular coordinates, or what convention you’re using for the names of the former? $(\rho,\theta,z)$ is at least as common for cylindrical coordinates as $(r,\theta,z)$. Don’t take for granted that others know what’s in your head: define your terms. $\endgroup$ – amd May 22 at 17:30
  • $\begingroup$ @amd You're being really silly here "please note I have to solve the problem particularly in the cylindrical coordinate system", secondly you know $x^2+y^2=r^2$ The Substitution is clear here. $\endgroup$ – EnlightenedFunky May 22 at 17:40
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The following integral setup solves for that particular problem is the following:\begin{equation} \int_0^{2\pi}\int_2^{2\sqrt{5}}\int_{-\sqrt{20-r^2}}^{\sqrt{20-r^2}}rdzdrd\theta+\int_0^{2\pi}\int_0^2\int_{-\sqrt{20-r^2}}^{r^2}rdzdrd\theta\end{equation} The result is the following which is \begin{equation}\iiint_Gf(r,\theta,z)dzdrd\theta=\frac{80\sqrt5+152}{3}\pi\end{equation}

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