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# Find the volume of the solid generated by revolving the standed region about the

Find the volume of the solid generated by revolving the standed region about the x-axis.

Te volume of the solid is ? cubic units.
The equation: $4x+3y=24$
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Philip O'Neill
Step 1
$⇒y=8-\left(\frac{4}{3}\right)x$
$⇒y=4×\left(2-\left(\frac{1}{3}\right)x\right)$
for x intercept, $y=0$
$0=4×\left(2-\left(\frac{1}{3}\right)x\right)$
$\left(\frac{1}{3}\right)x=2$
$x=6$
Volume generated by rotating the given region about x axis by washer method $v={\int }_{0}^{6}\pi {\left[4×\left(2-\left(\frac{1}{3}\right)x\right)\right]}^{2}dx$
$v={\int }_{0}^{6}16\pi \left[{2}^{2}-2×2\left(\frac{1}{3}\right)x+{\left(\left(\frac{1}{3}\right)x\right)}^{2}\right]dx$
$v={\int }_{0}^{6}16\pi \left[4-\left(\frac{4}{3}\right)x+\left(\frac{1}{9}\right){x}^{2}\right]dx$
$v={\int }_{0}^{6}16\pi \left[4x-\left(\frac{4}{3}\right)\left(\frac{1}{2}\right){x}^{2}+\left(\frac{1}{9}\right)\left(\frac{1}{3}\right){x}^{3}\right]$
$v={\int }_{0}^{6}16\pi \left[4x-\left(\frac{2}{3}\right){x}^{2}+\left(\frac{1}{27}\right){x}^{3}\right]$
$v=16\pi \left[4×6-\left(\frac{2}{3}\right){6}^{2}+\left(\frac{1}{27}\right){6}^{3}\right]-16\pi \left[4×0-\left(\frac{2}{3}\right){0}^{2}+\left(\frac{1}{27}\right){0}^{3}\right]$
$v=16\pi \left[4×6-\left(2×\frac{36}{3}\right)+\left(\frac{216}{27}\right)\right]$
$v=16\pi \left[24-24+8\right]$
$v=128\pi$
Volume $=128\pi =402$
###### Not exactly what you’re looking for?
Witheyesse47

Step 1
The disk formed when the skethched region revolves around the x-axis is shown below:

The given equation is, $4x+3y=24$
$3y=24-4x$
$y=8-\frac{4}{3}x$
To get x-intercept, equate $y=0$
$y=8-\frac{4}{3}x=0$
$8=\frac{4}{3}x$
$x=6$
Here, the disk radius $y=f\left(x\right)=8-\frac{4}{3}x$
And also x varies from 0 to 6.
Step 2
Volume of the solid of revolution obtained the given region about the x-axis is,
$V=\pi {\int }_{a}^{b}{\left[f\left(x\right)\right]}^{2}dx$
$=\pi {\int }_{0}^{6}{\left[8-\frac{4}{3}x\right]}^{2}dx$
$=\pi {\int }_{0}^{6}\left[64+\frac{16}{9}{x}^{2}-\frac{64}{3}x\right]dx$
$=\pi {\left[64x+\frac{64}{9}\left(\frac{{x}^{3}}{3}\right)-\frac{64}{3}\left(\frac{{x}^{2}}{2}\right)\right]}_{0}^{6}$
$=\pi \left[384+128-384\right]$
$=128\pi$
$=402.12$
Therefore the required volume of the solid is,
$=128\pi$ cubic units
$=402.12$ cubic units


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# How does one get time dilation, length contraction, and E=mc^2 from spacetime metric?

Tags:
1. Oct 20, 2014

### RelativeQuant

How does one get time dilation, length contraction, and E=mc^2 from the spacetime metric?

Suppose all that you are given is x12 + x22 + x32 - c2t2 = s2

How do you derive time dilation, length contraction, and E=mc^2 from this?

What is the most direct way to do this?

2. Oct 20, 2014

Staff Emeritus
You should take a look at Spacetime Physics, by Taylor and Wheeler, where this is all covered. It's only about 100 pages.

3. Oct 20, 2014

### RelativeQuant

Thanks! Is there a quick answer of any resources on the web? Thanks! :)

4. Oct 20, 2014

### RelativeQuant

5. Oct 20, 2014

### Staff: Mentor

Start with time dilation. The squared distance between (0,0,0,0) and (T,vT,0,0) must equal the squared distance between (0,0,0,0) and (T',0,0,0) because it's the same interval between the same two points, namely the endpoints of a journey taken at speed $v$ as viewed by the traveller (primed coordinates) and an observer moving at speed $v$ relative to the traveller (unprimed coordinates). Use the metric to calculate the squared distances, equate them, and solve for the ratio of T' to T.

6. Oct 20, 2014

### PAllen

This is not a simple. I don't think it can really be derived from the metric alone. You need to bring in, for example, energy and momentum conservation. Then, what you end up deriving from this plus the metric is E2 - p2c2 = m2c4 (once you have the right prior motivation, this just says that the 4-norm of m * 4-velocity unit vector is m, which is trivially true). When momentum is zero, you have the special case E=mc2.

Last edited: Oct 20, 2014


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# How does the temperature affect the amount of gas dissolved?

Feb 10, 2016

The solubility of gases in water increases when the temperature decreases.

#### Explanation:

The solubility of gases in water increases when the temperature decreases.

Why?

Gas particles have what is called an average kinetic energy $K {E}_{a v}$ which solely depends on temperature as follows:

$K {E}_{a v} = \frac{3}{2} R T$

Thus, when the temperature increases, the kinetic energy will increase and therefore, the gas particles will have enough energy to escape the liquid phase, and vice versa.


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Y-Delta conversion for Capacitors

cupid.callin
This was a ques in my book ... (Pic)

After thinking a while, i realized that it can be solved using Y-Delta (by converting Y to delta form) conversion. But i couldn't get the answer ... i don't know why ...

I used this http://en.wikipedia.org/wiki/Y-%CE%94_transform" [Broken]

It is given for resistors ... i guess its same for resistors and capacitors ... Right?

Please tell me if the eqn used is wrong or something else!

Attachments

• Y-Delta.jpg
25.4 KB · Views: 17,693
Last edited by a moderator:

Mentor
This was a ques in my book ... (Pic)

After thinking a while, i realized that it can be solved using Y-Delta (by converting Y to delta form) conversion. But i couldn't get the answer ... i don't know why ...

I used this http://en.wikipedia.org/wiki/Y-%CE%94_transform" [Broken]

It is given for resistors ... i guess its same for resistors and capacitors ... Right?

Please tell me if the eqn used is wrong or something else!

Remember how capacitors combined differently in series and parallel than do resistors?

What you can do is convert all your capacitances to their equivalent impedance, then use those formulae. Impedances mix and match like resistances.

Last edited by a moderator:
cupid.callin
How do i find impedance of a capacitor?

Mentor
How do i find impedance of a capacitor?

$$Zc = 1/(j\omega C$$)

$$\omega$$ is the operating frequency. The result is in Ohms, and will be an imaginary value.

Don't panic! You don't need to know the frequency for the math to work out; it's a constant for the given Y to Delta situation. If you do the algebra, a pretty simple result obtains. If the resistor version is:

Ra = (R1*R2 + R2*R3 + R3*R1)/R2
Rb = (R1*R2 + R2*R3 + R3*R1)/R3
Rc = (R1*R2 + R2*R3 + R3*R1)/R1

Then the capacitor version looks like:

Ca = C1*C3/(C1 + C2 + C3)
Cb = C1*C2/(C1 + C2 + C3)
Cc = C2*C3/(C1 + C2 + C3)

cupid.callin
And to find Y form,

C1 = (CaCb + CbCc + CcCa) / Cc
Right?

Thanks for the help gneill !
Thanks a lot !

Mentor
And to find Y form,

C1 = (CaCb + CbCc + CcCa) / Cc
Right?

Thanks for the help gneill !
Thanks a lot !

Right.

You're welcome.

Staff Emeritus
Homework Helper
Gold Member
... for the given Y to Delta situation. If you do the algebra, a pretty simple result obtains. If the resistor version is:

Ra = (R1*R2 + R2*R3 + R3*R1)/R2
Rb = (R1*R2 + R2*R3 + R3*R1)/R3
Rc = (R1*R2 + R2*R3 + R3*R1)/R1

Then the capacitor version looks like:

Ca = C1*C3/(C1 + C2 + C3)
Cb = C1*C2/(C1 + C2 + C3)
Cc = C2*C3/(C1 + C2 + C3)

The three capacitors enclosed in red form a Y. So does the other set of 1, 3, and 4 μF capacitors. Converting each of these sets to Δ configuration, as shown by gneill above, will allow you analyze the circuit as a combination of parallel and series capacitors.

cupid.callin
Thanks for the help!

PriyankB
Is there a simple result for delta to Y as well?

Mentor
Is there a simple result for delta to Y as well?

Sure. Just substitute the appropriate capacitor impedances into the formulas for resistance, stir and serve. Note that a capacitor impedance is of the form $1/(j \omega C)$.

So for example, given that for resistors
$$R_1 = \frac{R_a R_b}{R_a + R_b + R_c}$$
then
$$\frac{1}{C_1} = \frac{C_c}{C_b C_c + C_a C_c + C_a C_b}$$

and so on.

EDIT: Fixed up the expression. Should have been 1/C1 on the LHS.

Attachments

• Fig1.gif
2.3 KB · Views: 20,274
Last edited:
PriyankB
Thank you! :)


text:
Michael Cranston

UCI

## Time:

Wednesday, November 10, 2021 - 2:00pm to 3:00pm

## Location:

510R

In this talk we will discuss properties of integers selected according to the Riemann zeta distribution. We will emphasize two aspects of this distribution. The first is its faithful similarity to properties of an integer chosen according to the uniform distribution on a finite interval. The second aspect will be the appearance of Poisson behavior under this distribution. The Riemann zeta function is given for $\mbox{Re} z>1$ by

$$\zeta(z)=\sum_{n=1}^\infty \frac{1}{n^z}.$$

An alternative description is given by

$$\zeta(z)=\Pi_{p\in\mathcal{P}}\lt(1-\frac{1}{p^z}\rt)^{-1},$$

where $\mathcal{P}$ denotes the set of primes.

In our discussions we will replace the complex $z$ by a real number $s>1.$ We will denote by $X_s$ a random variable with the distribution

P(X_s=n)=\frac{1}{\zeta(s)n^s},\, n=1,2,3,\cdots.

The statistical properties of $X_s$ is the focus of the talk.

The talk is based on joint work with Adrien Peltzer.


text:
Substitution and Partial Fractions (Integration)

$$\int\frac{dx}{x-\sqrt[4]{x}}$$

given the substitution $x=u^{4}, dx=4u^{3}du$

$$=\int\frac{4u^{3}du}{u^{4}-u}=\int\frac{4u^{3}du}{u(u^{3}-1)}=\int\frac{4u^{2}du}{(u^{3}-1)}$$

At this point I believe I can factor the denominator and use partial fractions

$$\int\frac{4u^{2}du}{(u-1)(u^{2}+u+1)}=\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du$$

by substituting in $u=1$ and ignoring the factor in the denominator $(u-1)$ (Grubby Thumb Rule) $A=\frac{4*1^{2}}{1^{2}+1+1}, A=\frac{4}{3}$

and substituting arbitrarily $u=0$ into the expression $$4u^{2}=\frac{4}{3}(u^2+u+1)+B(u-1)$$ $$0=\frac{4}{3}(1)-B, B=\frac{4}{3}$$

therefore

$$\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du=\int\bigg(\frac{4}{3(u-1)}+\frac{4}{3(u^2+u+1)}\bigg)du$$

I am able to solve this however I dont think the value for B is correct

I have the following questions:

1: Am I able to use partial fractions even though the denominator of B is $u^{2}+u+1$

2: Did I do the substitution properly?

3: am I able to do what I did to find the value for A?

Thanks Stax

• Are you told that you have to do it with partial fractions? Because it's much easier without. Feb 5, 2016 at 1:00
• yes, I am supposed to use the provided substitution and then solve by partial fractions. I would be interested in the solution as you describe it though Feb 5, 2016 at 1:01
• I wonder about the wisdom of telling people to use partial fractions for a problem that can be done more easily by simpler methods. I posted such a method below. $\qquad$ Feb 5, 2016 at 1:09

If you have to use partial fractions, you need to start with $$\frac{4u^{2}}{(u-1)(u^{2}+u+1)}=\frac{A}{u-1}+\frac{Bu+C}{u^2+u+1}$$ and then follow pretty much the procedure you have tried.
The easy way to evaluate $$\int\frac{4u^{2}du}{u^{3}-1}=\frac43\int\frac{3u^{2}du}{u^{3}-1}$$ is by substituting $v=u^3-1$, which gives the answer immediately.
$$\int\frac{4u^{2}du}{u^3-1} = \frac 4 3 \int \frac {dw} w, \text{ etc.}$$