I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 12.12 : Cylindrical Coordinates
3. Convert the following equation written in Cartesian coordinates into an equation in Cylindrical coordinates.
\[{x^3} + 2{x^2} - 6z = 4 - 2{y^2}\]Show All Steps Hide All Steps
Start SolutionThere really isn’t a whole lot to do here. All we need to do is plug in the following \(x\) and \(y\) polar conversion formulas into the equation where (and if) possible.
\[x = r\cos \theta \hspace{0.5in}y = r\sin \theta \hspace{0.5in}{r^2} = {x^2} + {y^2}\] Show Step 2However, first we’ll do a little rewrite.
\[{x^3} + 2{x^2} + 2{y^2} - 6z = 4\hspace{0.5in} \to \hspace{0.5in}{x^3} + 2\left( {{x^2} + {y^2}} \right) - 6z = 4\] Show Step 3Now let’s use the formulas from Step 1 to convert the equation into Cylindrical coordinates.
\[{\left( {r\cos \theta } \right)^3} + 2\left( {{r^2}} \right) - 6z = 4\hspace{0.5in} \to \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{{r^3}{{\cos }^3}\theta + 2{r^2} - 6z = 4}}\]