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#### Chipset3600

##### Member

- Feb 14, 2012

- 79

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.

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- #1

- Feb 14, 2012

- 79

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.

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- #2

- Jan 26, 2012

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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?

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- #3

- Feb 14, 2012

- 79

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?

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- #4

- Jan 26, 2012

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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?

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- #5

- Feb 14, 2012

- 79

I prefer the first suggestion.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?

is correct my plot: http://i.imgur.com/0KfKPLG.jpg

actually i guess my best problem is in 3D graph =/

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- #6

- Jan 26, 2012

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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?

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- #7

- Feb 14, 2012

- 79

sorry didn't get the english, what is w.r.t?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?

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- #8

- Jan 26, 2012

- 4,205

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

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- #9

- Feb 14, 2012

- 79

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.My apologies: w.r.t. = "with respect to". It's a common phrase in calculus and analysis when taught in English.

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- #10

- Jan 26, 2012

- 4,205

Ok, let me use math instead. You have the following: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.

$$\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 $a,b$?

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- #11

- Feb 14, 2012

- 79

z= 0..3Ok, 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 $a,b$?

theta= 0..pi/2

r= 0..3 or the radius will not be ever 3 depending of angle?

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- #12

- Jan 26, 2012

- 4,205

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?

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- #13

- Feb 14, 2012

- 79

So the radius go from 0 to sqrt[(3-z^2)/(sin^2(theta))]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?

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- #14

- Jan 26, 2012

- 4,205

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

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- #15

- Feb 14, 2012

- 79

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

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- #16

- Jan 26, 2012

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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?Can you show me the graph of bounded area please?

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- #17

- Feb 14, 2012

- 79

from tan^-1(3) to pi/2?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?

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- #18

- Jan 26, 2012

- 4,205

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

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- #19

- Feb 14, 2012

- 79

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

Last edited:

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- #20

- Jan 26, 2012

- 4,205

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?

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- #21

- Jan 26, 2012

- 4,205

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?

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- #22

- Feb 14, 2012

- 79

Calculate the ∭zdVHmm. Not sure I can finish, actually. The $\theta$ integral doesn't converge.

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

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.

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- #23

- Feb 7, 2012

- 2,818

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: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.

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.

- Jan 26, 2012

- 183

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- #25

- Jan 26, 2012

- 4,205

$$\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.