# Cylindrical coordinates on elliptic paraboloids.

I want to compute the volume bounded by:

• the cylinder $x^2+4y^2=4$.
• the $z=0$ plane.
• the elliptic paraboloid $z = x^2 + 6y^2$.

I would like to use cylindrical coordinates. However I have never dealt with a problem in which $r$ bounds depend on $\theta$.

Here there's a horrible sketch of the solid: The base is an ellipse. So I take the transformation to cylindrical coordinates: \begin{align} x&=r\cos\theta,\\ y&=r\sin\theta, \\ z&=z. \end{align} Which implies $0\leq z\leq x^2+6y^2 = r^2\cos^2\theta + 6r^2\sin^2\theta$ and $0\leq\theta\leq2\pi$.

However I'm having difficulties in determining the $r$ bounds. Also I would like to know the equation of the curve of intersection between the cylinder and the paraboloid, but don't know how to do it.

For the $r$ bounds I tried considering the equation for the base: $$4y^2+x^2=4\Rightarrow 4r^2\sin^2\theta+r^2\cos^2\theta = 4 \Rightarrow r=\frac{2}{\sqrt{4\sin^2\theta+\cos^2\theta}}.$$

I'm not sure if doing that is correct, also the integral looks absolutely grim. I really appreciate your help.

• I think this web page might be useful for you, math.byu.edu/~math302/content/learningmod/coordinates/… at the end, there are all the integration orders from cylindrical coordinates. – Keith Nov 10 '14 at 1:56
• Rather than use cylindrical coordinates immediately, it may be easier to make a $x=2u$ substitution and work with cylindrical coordinates in this $(u,y,z)$ coordinate system. – Semiclassical Nov 10 '14 at 2:27

You are given equations in Cartesian coordinates. At first you need to seek intersection in the same system for handling convenience of given equations. Later on, go in for cylindrical coordinates. But do not a priori choose to adopt cylindrical coordinates.

To handle the total situation, choose a proper parameter arising out of given Cartesian equations like u rather than assume that cylindrical $\theta$ could be the needed parameter.

Eliminate y between cylinder and elliptic paraboloid equations:

$$x^2 + 4 y^2 = 4, z –x^2 – 6 y^2 = 0$$ to get $$2 z + x^2 = 12$$

Its parametric coordinates are:

$$z= 6 \cos^2 u , x = 2 \sqrt{3} \ sin\, u.$$

Eliminate x between the same equations to get

$$z \, – 2\, y^2 = 4$$ giving $$y= \pm \sqrt {(3 cos^2 u -2)}$$

You can see the demarcating ring between elliptic cylinder and elliptic paraboloid. I have thickened this line as a tube to visualize, as it is the central object of your query.

$$x = 2 \sqrt{3} \sin u, y= \pm \sqrt{( 3 cos^2 u -2)} , z= 6 \cos^2 u, r= 1 + 9\sin^2\,u$$

Now, you can convert parametric equations $z,r, u$ to $r$ and $\theta \,$ your preferred cylindrical coordinates, using $tan^2 \theta = \frac {( 3 cos^2 u -2)} { 12 \sin ^2 u }$, $dV= 4 \,r\,dz\,d\theta\,dr$ etc. However, it is advantageous with u as it is, to integrate between limits $u = 0,2 \pi$ :

Mathematica program:

x[u_]=2 Sqrt Sin[u];y[u_]=Sqrt[3 Cos[u]^2-2];z[u_]=6Cos[u]^2;
aa=ParametricPlot3D[{x[u],-y[u],z[u]},{u,0,2 Pi},PlotStyle->{Thick,Tube[.07]}];
bb=ParametricPlot3D[{x[u],y[u],z[u]},{u,0,2 Pi},PlotStyle->{Thick,Tube[.07]}];
Show[{aa,bb},PlotRange->All]
EllCyl=ParametricPlot3D[{2 Cos[u],Sin[u],v},{u,0,2 Pi},{v,-0,6.5},PlotStyle->{Green,Opacity[0.5]}]
EllPar=ContourPlot3D[z-x^2-6 y^2==0,{x,-3,3},{y,-2,2},{z,0,6.5}]
Show[{aa,bb,EllCyl,EllPar},PlotRange->All]
Plot[{x[u],y[u],z[u]},{u,0,2 Pi},PlotStyle->{Thick,Red},GridLines->Automatic]
Plot[y[u],{u,0,2 Pi},PlotStyle->{Thick,Blue},GridLines->Automatic]
th[u_]:=ArcTan[x[u],y[u]]
ParametricPlot[{th[u],Sqrt[x[u]^2+y[u]^2]},{u,0,2 Pi},GridLines->Automatic]
ParametricPlot[{{th[u],x[u]},{th[u],y[u]},{th[u]+Pi,-y[u]},{th[u],z[u]}},{u,0,2 Pi},GridLines->Automatic,PlotStyle->{Thick,Magenta}]
Plot[th[u],{u,0,2 Pi},GridLines->Automatic]
" CYL COORDS ON ELLIPTIC  PARABOLOIDS   SE NOV 2014"


You can appreciate the relationship between $r, z$ and $\theta$ using chosen parameter $u$, if you dont want to consider change of independent variable to $u$. I indicated a graph for $\theta, r$ of demarcation line in the above sketch.

Consider the change of variables $x=2u\Rightarrow dx = 2\,du$. Now we are working on a circular cylinder: $u^2+y^2=1$. Now let's take cylindric coordinates. Note that the Jacobian is $2r$. And the volume: \begin{align} V=\int_0^1\int_0^{2\pi}\int_0^{4r^2\cos^2\theta+6r^2\sin^2\theta} 2r\,dz\,d\theta\,dr = \int_0^1\int_0^{2\pi} 8r^3\cos^2\theta+12r^3\sin^2\theta\,d\theta\,dr\\ =\int_0^1\int_0^{2\pi} \overbrace{ 4r^3+4r^3\cos2\theta+6r^3-6r^3\cos2\theta}^{2\theta = s\;\Rightarrow\; 2\,d\theta=ds} \,d\theta\,dr\\ =\int_0^2\left[8\pi r^3+\int_0^{4\pi}2r^3\cos s\,ds+12\pi r^3 - \int_0^{4\pi}3r^3\cos s\,ds\right]dr\\ =\int_0^1 20\pi r^3\, dr = 5\pi. \end{align}