Help with center of mass question

[student93]

Homework Statement

The problem is attached in this post.

Homework Equations

x=My/M=(∫xf(x)dx)/(∫f(x)dx) from a to b

The Attempt at a Solution

My=4∫xtan(x^2)dx from 0 to √π/2 = 2ln(2/√2)

M=4∫tan(x^2)dx from 0 to √π/2 =2xln(2/√2)

My/M= 2ln(2√2)/2xln(2/√2) = 1/x or 1/A

The answer however is (1/A)ln2

 

[Adithyan]

Homework Statement

The problem is attached in this post.

Homework Equations

x=My/M=(∫xf(x)dx)/(∫f(x)dx) from a to b

The Attempt at a Solution

My=4∫xtan(x^2)dx from 0 to √π/2 = 2ln(2/√2)

Wrong.

You have calculated the area but ∫xdm or 'My' (as you have put it) is
$σ*∫xf(x)dx$ {σ = Mass per unit Area}
$∴ My = σ*2ln√2$

M=4∫tan(x^2)dx from 0 to √π/2 =2xln(2/√2)

Wrong.
First of all, you have calculated the area incorrectly. This integration is different from the previous one- there is no 'x' over here.
Moreover,
M = ∫dm = ∫σdA (Where σ= Mass per unit Area)=σ*Area.


Hope you got it!
Adithyan.

 

[student93]

For "My", I'm still getting -2ln(√2/2) as the value of my integral. How did you get 2ln√2? Could you please clarify how you did the integration etc.?

 

[Adithyan]

For "My", I'm still getting -2ln(√2/2) as the value of my integral. How did you get 2ln√2? Could you please clarify how you did the integration etc.?

$-2ln(√2/2) = -2ln(1/√2) = -2ln(√2^{-1}) = (-2)*(-1)ln(√2)= 2ln√2.$

I guess, you are okay with my answer now. By the way, I did the first integration (ie $∫xtan(x^{2})dx$) by putting $x^{2} = t$. (I hope you are well versed with integration by substitution.)

 

[student93]

Yeah I also used substitution, also could you also please clarify the integration for "M", I keep getting -2xln(√2/2), which is equal to 2xln(ln√2) (Is this correct?). Also what exactly do I do with the "x" in my answer? (I got the "x" because I used substitution to solve for this integral).

Also the answer to the question is (ln2)/A, where "A" is equal to Area, however I don't know how to apply the variable "A" in regards to this question.

 

[Adithyan]

Yeah I also used substitution, also could you also please clarify the integration for "M", I keep getting -2xln(√2/2), which is equal to 2xln(ln√2) (Is this correct?). Also what exactly do I do with the "x" in my answer? (I got the "x" because I used substitution to solve for this integral).

Also the answer to the question is (ln2)/A, where "A" is equal to Area, however I don't know how to apply the variable "A" in regards to this question.

Let
$σ =\frac{Mass}{Area }.$


$M = σ*A$

$∴dm = σdA$
$M = ∫dm = ∫σdA =σ∫dA = σ*A$

Here $A =∫f(x)dx = ∫tan(x^{2})dx$
and in this integration you cannot use substitution method since, there is no xdx term in it.

 

[student93]

Is it possible to integrate 4tan(x^2) then? How exactly do you go about integrating a function such as 4tan(x^2) etc.?

 

[Adithyan]

I tried some of the common methods of integration, but they get messy later on. I don't think I can integrate this term at my current level of understanding of integration. Moreover, the question didn't ask for the value, they themselves substituted the value of this integral with A.

 

[student93]

Ok that makes sense since I also had a lot of trouble trying to integrate 4tan(x^2).

Also after following the steps you provided, I ended up getting my answer to be (2ln2)/A instead of the actual answer (ln2)/A

I set x=My/M=((p)(2ln2))/((p)(A))

My=(p)(2ln2)
M=(p)(A)

p=density
A=Area

Did I make an error in my setup?

 

[Adithyan]

Also after following the steps you provided, I ended up getting my answer to be (2ln2)/A instead of the actual answer (ln2)/A

I set x=My/M=((p)(2ln2))/((p)(A))

My=(p)(2ln2)
M=(p)(A)

p=density
A=Area

Did I make an error in my setup?

My=(p)(2ln2)

You made an error here.
You should get this term after integration: $2∫^{\frac{√\pi}{2}}_{0}ln(sec(x^{2}))$

 

[student93]

I don't understand how you got that as the integral, after substitution, don't the limits of integration become π/4 and 0?

My=(p)4∫xtan(x^2)dx from 0 to √π/2

My=(p)4∫xtan(x^2)dx from 0 to √π/2

u=x^2 -> 0 to π/4 become the new limits of integration
du=2xdx

My=(p)(4/2)∫tan(u)du from 0 to π/4

My=(p)(2)(-ln(cos(u)) from 0 to π/4

My=(p)((-2ln(√2/2)-(-2ln(1))

My=(p)(-2ln(√2/2))=2ln(2)

 

[Adithyan]

I don't understand how you got that as the integral, after substitution, don't the limits of integration become π/4 and 0?

My=(p)4∫xtan(x^2)dx from 0 to √π/2

My=(p)4∫xtan(x^2)dx from 0 to √π/2

u=x^2 -> 0 to π/4 become the new limits of integration
du=2xdx

My=(p)(4/2)∫tan(u)du from 0 to π/4

My=(p)(2)(-ln(cos(u)) from 0 to π/4

My=(p)((-2ln(√2/2)-(-2ln(1))

My=(p)(-2ln(√2/2))=2ln(2)

I think you need to get your logarithmic knowledge correct.
$-2(ln(√2/2)) = -2(ln(\frac{1}{√2}) = 2ln(√2)= ln(√2^{2}) = ln(2)$

Here, go through these rules and learn them to avoid such calculation errors in future.

http://en.wikipedia.org/wiki/List_of_logarithmic_identities