Converting integration of rectangular integral to spherical.

I don't think it's possible. I don't think I can set the upper bound of ##\rho## in such a way that I can have the whole region within the constraints for ##\theta## and ##\phi##.
  • #1
Eclair_de_XII
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Homework Statement
\begin{eqnarray}
0\leq z\leq \sqrt{x^2+y^2}\\
0\leq y\leq x\\
0\leq x\leq 1
\end{eqnarray}

##f(x,y,z)=\frac{(x^2+y^2)^\frac{3}{2}}{x^2+y^2+z^2}##
Relevant Equations
\begin{eqnarray}
x=\rho\cos\theta\cos\phi\\
y=\rho\sin\theta\cos\phi\\
z=\rho\sin\phi
\end{eqnarray}

Answer given in book: ##\frac{\pi}{12}##
Answer derived: ##\frac{3}{16}{\pi}##
I'm going to type out my LaTeX solution later on. But in the meantime, can anyone check my work? I know it's sloppy, disorganized, and skips far more steps than I care to count, but I'd very much appreciate it. I'm not getting the answer as given in the book. I think I failed this time because I assumed that ##\phi## was the angle from the x-axis (or any other axis that can be constructed with a line on the xy-plane) to some arbitrary point on the zx-plane. When I retried it while treating ##\phi## as the angle from the z-axis to some arbitrary point on the zx-plane, I got ##\frac{3}{4}\pi(\frac{\pi}{8}-1)##. Can anyone point out anything I might have done wrong in the photograph I posted while I type out a LaTeX version of my second attempt? Thank you.
 

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  • #2
1 Convert rectangular bounds of integration into spherical bounds.
2 Convert ##f## to spherical form.
3 Integrate.

===1===

\begin{eqnarray}
x=\rho\cos\theta\sin\phi\\
y=\rho\sin\theta\sin\phi\\
z=\rho\cos\phi
\end{eqnarray}

\begin{eqnarray}
0\leq z\leq \sqrt{x^2+y^2}\\
0\leq y\leq x\\
0\leq x\leq 1
\end{eqnarray}

\begin{eqnarray}
x^2+y^2&=&(\rho\cos\theta\sin\phi)^2+(\rho\sin\theta\sin\phi)^2\\
&=&(\rho^2\cos^2\theta\sin^2\phi+\rho^2\sin^2\theta\sin^2\phi)\\
&=&\rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta)\\
&=&\rho^2\sin^2\phi
\end{eqnarray}

\begin{eqnarray}
z=\sqrt{x^2+y^2}\\
\rho\cos\phi=\rho|\sin\phi|\\
\rho(\cos\phi-|\sin\phi|)=0
\end{eqnarray}

We're investigating a region in the first octant, so ##\sin\phi>0##, and so ##\phi=\frac{\pi}{4}##.

\begin{eqnarray}
z=0\\
\rho\cos\phi=0
\end{eqnarray}

Hence, ##\phi=\frac{\pi}{2}##.

\begin{eqnarray}
y=x\\
\rho\sin\theta\sin\phi=\rho\cos\theta\sin\phi\\
\rho\sin\phi(\sin\theta-\cos\theta)=0
\end{eqnarray}

We are investigating the projection of the curve onto the xy-plane, so ##\phi=\frac{\pi}{2}##, and so, ##\theta=\frac{\pi}{4}##.

\begin{eqnarray}
y=0\\
\rho\sin\theta\sin\phi=0
\end{eqnarray}

Similarly, ##\theta=0##.

It is given that ##x\leq1##, and that ##y\leq1##.
It is given also, that ##z\leq\sqrt{x^2+y^2}\leq\sqrt{1^2+1^2}\leq\sqrt{2}##.

Hence, ##x^2+y^2+z^2\leq1+1+2=4=2^2##, and so ##\rho\leq2##.

Our spherical bounds are therefore:

\begin{eqnarray}
0\leq\rho\leq2\\
\frac{\pi}{4}\leq\phi\leq\frac{\pi}{2}\\
0\leq\theta\leq\frac{\pi}{4}
\end{eqnarray}

===2===

\begin{eqnarray}
f&=&\frac{(x^2+y^2)^{\frac{3}{2}}}{x^2+y^2+z^2}\\
&=&\frac{\rho^3\sin^3\phi}{\rho^2}\\
&=&\rho\sin^3\phi
\end{eqnarray}

===3===

\begin{eqnarray*}
\int_0^{\frac{\pi}{4}}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_0^2(\rho\sin^3\phi)(\rho^2\sin\phi) d\rho d\phi d\theta&=&\int_0^{\frac{\pi}{4}}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_0^2\rho^3\sin^4\phi d\rho d\phi d\theta\\
&=&\frac{1}{4}\rho^4|_0^2\cdot\frac{\pi}{4}\cdot\int (1-\cos^2\phi)^2 d\phi\\
&=&\frac{1}{4}2^4\cdot\frac{\pi}{4}\cdot\int(1-2\cos^2\phi+\cos^4\phi) d\phi\\
&=&\frac{1}{4}\cdot16\cdot\frac{\pi}{4}\cdot\left[\phi-\int(1+\cos2\phi) d\phi+\frac{1}{4}\int(1+\cos2\phi)^2 d\phi\right]\\
&=&4\cdot\frac{\pi}{4}\cdot[\phi-\phi-\sin2\phi+\frac{1}{4}\int(1+2\cos2\phi+\cos^2(2\phi))^2 d\phi]\\
&=&\pi\cdot\left[-\sin2\phi+\frac{1}{4}(\phi+\sin2\phi+\frac{1}{2}\phi+\frac{1}{4}\sin4\phi)\right]\\
&=&\pi\cdot\left[-\sin2\phi+\frac{1}{4}(\phi+\sin2\phi+\frac{1}{2}\phi+\frac{1}{4}\sin4\phi)\right]\\
&=&\pi\left[\frac{3}{8}\phi-\frac{3}{4}\sin2\phi+\frac{1}{16}\sin4\phi\right]|_{\frac{\pi}{4}}^{\frac{\pi}{2}}\\
&=&\frac{3}{4}\pi(\frac{\pi}{8}-1)
\end{eqnarray*}
 
Last edited:
  • #4
One thing that confuses me is that you use ##\theta## for the azimuth angle and ##\phi## for the inclination angle, while I am used the other way around, but ok I ll try to overcome this.
 
  • #5
Section 2 looks correct
Section 3 doesn't look correct but even after I do the correction (wolfram says that $$\int \sin^4\phi d\phi=\frac{3}{8}\phi-\frac{1}{4}\sin2\phi+\frac{1}{32}\sin4\phi$$
i don't get the book answer.

SO you big mistake must be in correctly determining the spherical boundaries.
 
  • #6
Okay, gotcha. I'll work on it. I very much suspect it's got something to do with the ##\rho## upper-bound.

It's got to rely on ##\theta## and ##\phi##, I think.
 
Last edited:
  • #7
Okay, so it turns out that this problem probably wanted me to use cylindrical coordinates, instead.

I cannot fathom how it can be done with spherical coordinates.
 

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  • #8
Good that you did it in cylindrical coordinates and confirmed the book answer. (Cant check you work though cause no matter how much I tried to magnify it looks faint , I just can't make out all the symbols.)

When you find time post your work with cylindrical coordinates in Latex please.
 
  • #9
I found the bounds for ##r## by looking at the projection of the surface onto the xy-plane. (See illustration in OP).

===1: Converting to cylindrical coordinates===

\begin{eqnarray}
x=1\\
r\cos\theta=1\\
r=\sec\theta
\end{eqnarray}

\begin{eqnarray}
0\leq r\leq \sec\theta\\
0\leq\theta\leq\frac{\pi}{4}\\
0\leq z\leq r
\end{eqnarray}

\begin{eqnarray}
f&=&\frac{(x^2+y^2)^\frac{3}{2}}{x^2+y^2+z^2}\\
&=&\frac{r^3}{r^2+z^2}
\end{eqnarray}

===2: Solving the integral===

\begin{eqnarray}
z=r\tan s\\
dz=r\sec^2 s\,ds\\
z^2+r^2=r^2\sec^2s\\
s&:&r\mapsto\frac{\pi}{4}\\
&:&0\mapsto0\\
\end{eqnarray}

\begin{eqnarray}
\int_0^\frac{\pi}{4}\int_0^{\sec\theta}\int_0^\frac{\pi}{4}\left(\frac{r^3}{r^2\sec^2s}\right)r(r\sec^2s\,ds)\,dr\,d\theta&=&\int_0^\frac{\pi}{4}\int_0^{\sec\theta}\int_0^\frac{\pi}{4}r^3\,ds\,dr\,d\theta\\
&=&s|_0^\frac{\pi}{4}\cdot\int_0^\frac{\pi}{4}\int_0^{\sec\theta}r^3\,dr\,d\theta\\
&=&\frac{\pi}{4}\cdot\frac{1}{4}\int_0^\frac{\pi}{4}\sec^4\theta\,d\theta\\
&=&\frac{\pi}{16}\cdot\int_0^\frac{\pi}{4}(1+\tan^2\theta)\sec^2\theta\,d\theta\\
&=&\frac{\pi}{16}\cdot\left(\tan\theta+\frac{1}{3}\tan^3\theta\right)|_0^\frac{\pi}{4}\\
&=&\frac{\pi}{16}\cdot\left(1+\frac{1}{3}\right)\\
&=&\frac{\pi}{16}\cdot\left(\frac{4}{3}\right)\\
&=&\frac{\pi}{12}
\end{eqnarray}

And I think it could be possible to solve this in spherical coordinates, if you set the upper bound of ##\rho=\sec\theta\csc\phi##.
 
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  • #10
Yes well this seems correct, initially I had my doubts about the boundaries for ##\theta## but after looking through some inequalities at wolfram (namely ##0<\sin\theta<\cos\theta##) I think they are right.

I am not sure at all about the boundaries in spherical coordinates, to tell you the truth I am not so good in converting 3D domains between cartesian/spherical/cylindrical coordinates systems , that's why in my first post I checked first section 2 and 3 of yours.
 

1. What is the purpose of converting rectangular integration to spherical?

The purpose of converting rectangular integration to spherical is to simplify the integration process and make it easier to solve certain types of integrals. It is also useful for finding solutions in a different coordinate system, such as spherical coordinates.

2. How do you convert from rectangular to spherical coordinates?

To convert from rectangular to spherical coordinates, you can use the following equations:

x = r sin(θ) cos(φ)

y = r sin(θ) sin(φ)

z = r cos(θ)

where r is the distance from the origin, θ is the angle from the positive z-axis, and φ is the angle from the positive x-axis to the projection of the point onto the xy-plane.

3. What types of integrals can be converted from rectangular to spherical?

Integrals that involve spherical symmetry or involve functions that are easier to integrate in spherical coordinates can be converted from rectangular to spherical. Examples include integrals involving spheres, cones, and cylinders.

4. Are there any limitations to converting from rectangular to spherical coordinates?

Yes, there are some limitations to converting from rectangular to spherical coordinates. This method may not be suitable for all types of integrals and may not always result in a simpler solution. Additionally, the conversion may not be possible for certain regions of integration.

5. Can converting to spherical coordinates change the bounds of integration?

Yes, converting to spherical coordinates can change the bounds of integration. This is because the new coordinate system may have a different shape and orientation compared to the original rectangular coordinates. It is important to carefully consider the new bounds of integration when converting to spherical coordinates.

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