# Integral - cylindrical coordinates

#### Chipset3600

##### Member
Hello, my best problem is about find the integration limits.
in cylindrical coordinates- where V is limited by the cylinder y^2+z^2=9 and the planes x = 0, y = 3x and z = 0 in the first octant.

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

So, let's say that you have
\begin{align*}
z&=z\\
y&=r \sin(\theta) \\
x &= r \cos(\theta).
\end{align*}

You have several constraints. How about being inside the cylinder $y^{2}+z^{2}=9$? How about being in the first octant? How about being bound by the three planes? How could you restrict your three integration variables, $z,r,\theta$, to account for these constraints?

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

So, let's say that you have
\begin{align*}
z&=z\\
y&=r \sin(\theta) \\
x &= r \cos(\theta).
\end{align*}

z=0..3
y= 0..sqrt(9-y^2)
x=0..y/3
it is correct?

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

Hmm. I'm not sure I agree with that.

Step one: determine in what order you're going to integrate. I'd recommend either
$$\iiint r\, dr\, d\theta\, dz$$
or
$$\iiint dz\, r\, dr\, d\theta.$$

Step two: each integral can have variables in the integration limits that appear outside the integral, but not inside. So you could have
$$\int_{0}^{1} \int_{0}^{z} \int_{0}^{\theta}r\, dr\, d\theta\, dz,$$
but not
$$\int_{0}^{\theta} \int_{0}^{z} \int_{0}^{1} r\, dr\, d\theta\, dz.$$
So what constraints does that place on you?

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

Hmm. I'm not sure I agree with that.

Step one: determine in what order you're going to integrate. I'd recommend either
$$\iiint r\, dr\, d\theta\, dz$$
or
$$\iiint dz\, r\, dr\, d\theta.$$

Step two: each integral can have variables in the integration limits that appear outside the integral, but not inside. So you could have
$$\int_{0}^{1} \int_{0}^{z} \int_{0}^{\theta}r\, dr\, d\theta\, dz,$$
but not
$$\int_{0}^{\theta} \int_{0}^{z} \int_{0}^{1} r\, dr\, d\theta\, dz.$$
So what constraints does that place on you?
I prefer the first suggestion.
is correct my plot: http://i.imgur.com/0KfKPLG.jpg
actually i guess my best problem is in 3D graph =/

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

Not too bad. You also have a plane coming out from the $z$-axis, $y=3x$.

So if you're integrating w.r.t. $r$ first, what is it going to range from? You're allowed to have $r=r(\theta,z)$. Then you integrate w.r.t. $\theta$, and you're allowed to have $\theta=\theta(z)$. Finally, you integrate w.r.t. $z$, and $z$ must range from one number to another. What will you have?

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

Not too bad. You also have a plane coming out from the $z$-axis, $y=3x$.

So if you're integrating w.r.t. $r$ first, what is it going to range from? You're allowed to have $r=r(\theta,z)$. Then you integrate w.r.t. $\theta$, and you're allowed to have $\theta=\theta(z)$. Finally, you integrate w.r.t. $z$, and $z$ must range from one number to another. What will you have?
sorry didn't get the english, what is w.r.t?

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

sorry didn't get the english, what is w.r.t?
My apologies: w.r.t. = "with respect to". It's a common phrase in calculus and analysis when taught in English.

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

My apologies: w.r.t. = "with respect to". It's a common phrase in calculus and analysis when taught in English.
Sorry but I am not able to follow the reasoning. Maybe the technical english dont help me, i just tried understand something in forum just interpreting the equations.

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

Sorry but I am not able to follow the reasoning. Maybe the technical english dont help me, i just tried understand something in forum just interpreting the equations.
Ok, let me use math instead. You have the following:
$$\int_{a}^{b} \int_{\theta_{1}(z)}^{\theta_{2}(z)} \int_{r_{1}(\theta,z)}^{r_{2}( \theta,z)} r \, dr\, d\theta \, dz.$$
What are the functions $r_{1},r_{2},\theta_{1},\theta_{2}$ and the constants $a,b$?

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

Ok, let me use math instead. You have the following:
$$\int_{a}^{b} \int_{\theta_{1}(z)}^{\theta_{2}(z)} \int_{r_{1}(\theta,z)}^{r_{2}( \theta,z)} r \, dr\, d\theta \, dz.$$
What are the functions $r_{1},r_{2},\theta_{1},\theta_{2}$ and the constants $a,b$?
z= 0..3
theta= 0..pi/2
r= 0..3 or the radius will not be ever 3 depending of angle?

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

I don't think you've quite got it yet. Your $r$ is going to have to depend on either $\theta$ or $z$ or both. You know that $r^{2}=x^{2}+y^{2}$, and that $\theta=\tan^{-1}(y/x)$. Can you use these relationships to find an upper bound for $r$? I would agree with the lower bound: $0<r$.

Try this: $y^{2}+z^{2}=3$, so $x^{2}+y^{2}+z^{2}=3+x^{2}$, and thus $r^{2}+z^{2}=3+x^{2}=3+r^{2}\cos^{2}(\theta)$. Or, more directly, you can say that since $y=r \sin(\theta)$, that $r^{2} \sin^{2}(\theta)+z^{2}=3$. Solving for $r^{2}$ yields that
$$r^{2}= \frac{3-z^{2}}{ \sin^{2}( \theta)}.$$
Can you continue from here?

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

I don't think you've quite got it yet. Your $r$ is going to have to depend on either $\theta$ or $z$ or both. You know that $r^{2}=x^{2}+y^{2}$, and that $\theta=\tan^{-1}(y/x)$. Can you use these relationships to find an upper bound for $r$? I would agree with the lower bound: $0<r$.

Try this: $y^{2}+z^{2}=3$, so $x^{2}+y^{2}+z^{2}=3+x^{2}$, and thus $r^{2}+z^{2}=3+x^{2}=3+r^{2}\cos^{2}(\theta)$. Or, more directly, you can say that since $y=r \sin(\theta)$, that $r^{2} \sin^{2}(\theta)+z^{2}=3$. Solving for $r^{2}$ yields that
$$r^{2}= \frac{3-z^{2}}{ \sin^{2}( \theta)}.$$
Can you continue from here?
So the radius go from 0 to sqrt[(3-z^2)/(sin^2(theta))]

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

So the radius go from 0 to sqrt[(3-z^2)/(sin^2(theta))]
Right. Now, what are the limits on $\theta$? Hint: it'll be partly determined by the plane $y=3x$.

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

Right. Now, what are the limits on $\theta$? Hint: it'll be partly determined by the plane $y=3x$.
Can you show me the graph of bounded area please?

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

Can you show me the graph of bounded area please?
Alas, I do not have very good graphing or scanning capabilities where I am. The plane $y=3x$ does not depend on $z$, so it'll look the same no matter what $z$ is. It will intersect with the $z$-axis, and come out of it at an angle with respect to the $x$-axis. Can you find what that angle is?

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

Alas, I do not have very good graphing or scanning capabilities where I am. The plane $y=3x$ does not depend on $z$, so it'll look the same no matter what $z$ is. It will intersect with the $z$-axis, and come out of it at an angle with respect to the $x$-axis. Can you find what that angle is?
from tan^-1(3) to pi/2?

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

Close. I think it's $0$ to $\tan^{-1}(3)$.

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

Close. I think it's $0$ to $\tan^{-1}(3)$.
So the integral be: $$\displaystyle \int_{0}^{3}\int_{0}^{\arctan 3 }\int_{0}^{\sqrt{\frac{3-z^2}{sin^2(\theta)}}} rdr d\theta dz$$ ?? this equal to Zero :s

Last edited:

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

I don't think the integral will be zero. You have
\begin{align*}
\int_{0}^{3} \int_{0}^{\tan^{-1}(3)} \int_{0}^{\frac{\sqrt{3-z^{2}}}{\sin(\theta)}}
r\, dr\, d\theta\, dz&=
\int_{0}^{3} \int_{0}^{ \tan^{-1}(3)} \left( \frac{r^{2}}{2} \right) \Bigg|_{0}^{ \frac{ \sqrt{3-z^{2}}}{ \sin( \theta)}}
d\theta\, dz\\
&=\frac{1}{2}\int_{0}^{3} \int_{0}^{ \tan^{-1}(3)} \frac{3-z^{2}}{\sin^{2}( \theta)}
d\theta\, dz\\
&= \frac{1}{2} \left( \int_{0}^{3}(3-z^{2})\,dz \right) \left( \int_{0}^{\tan^{-1}(3)} \csc^{2}( \theta) \, d\theta \right).
\end{align*}
Can you finish?

#### Ackbach

##### Indicium Physicus
Staff member
Re: Integral- cylindrical coordinates

Hmm. Not sure I can finish, actually. The $\theta$ integral doesn't converge.

Can you write down the original problem statement, word-for-word, please?

#### Chipset3600

##### Member
Re: Integral- cylindrical coordinates

Hmm. Not sure I can finish, actually. The $\theta$ integral doesn't converge.

Can you write down the original problem statement, word-for-word, please?
Calculate the ∭zdV
In cylindrical coordinates: where V is limited by the cylinder y^2+z^2=9 and the planes x = 0, y = 3x and z = 0 in the first octant.

#### Opalg

##### MHB Oldtimer
Staff member
Re: Integral- cylindrical coordinates

Calculate the ∭zdV
In cylindrical coordinates: where V is limited by the cylinder y^2+z^2=9 and the planes x = 0, y = 3x and z = 0 in the first octant.
I have kept away from this discussion because I could not see a convenient way of solving the problem using cylindrical coordinates. In fact, I think it is much better not to use them. Suppose instead that we look at a cross-section of $V$ at a fixed level of $z$. The cross-section looks like this:

We must have $y$ lying between $-\sqrt{9-z^2}$ and $+\sqrt{9-z^2}$. We also need $x\geqslant0$. Next, the line $y=3x$ must mark the boundary of the cross-section. But which side of that line should $V$ be on? The wording of the problem does not make that clear, and that seems to be one of the reasons that this problem is causing so much discussion. But if $V$ is to be bounded, it is clear from the picture that it must lie to the left of the line, so that the cross-section is the pink shaded region in the diagram. This is a triangle whose height is $\sqrt{9-z^2}$ and whose width is one-third of that. Thus its area is $\frac16(9-z^2)$. Finally, since $V$ lies in the positive octant, we must have $0\leqslant z\leqslant 3$. It follows that $$\displaystyle \iiint z\,dV = \int_0^3 \tfrac16z(9-z^2)\,dz,$$ which is an easy integral.​

#### Jester

##### Well-known member
MHB Math Helper
Here's a 3D picture. Surfaces in 3D are hard for a lot of people to visualize. For me, I'd stick with Cartesian coords or go with a different parameterization than cylindrical polars. (Sorry it's so large - maybe someone can scale it down abit.

#### Ackbach

##### Indicium Physicus
Staff member
Thanks much for the corrections from Opalg and the picture from Jester. So, if you must do the integral in cylindrical, the appropriate integral is
$$\int_{0}^{3} \int_{\tan^{-1}(3)}^{ \frac{ \pi}{2}} \int_{0}^{\frac{\sqrt{3-z^{2}}}{\sin(\theta)}} r\, dr\, d\theta\, dz.$$
This time, the $\csc^{2}$ integral will converge.