Surface area of a circular cylinder cut by a slanted plane

[LaplacianHarmonic]

Homework Statement

Cylinder : x^2 + y^2 = 1
Plane that intersects above cylinder: y + z = 2

What is the surface area of the sides of this cylinder?

Homework Equations

dS= R1 d@ dz
@ is from 0 to 2 pi
z is from 0 to 2 - y

dS=(Zx^2 + Zy^2 + 1)^.5 dA
Where Z = 2 - y

The Attempt at a Solution

I used
dS=(Zx^2 + Zy^2 + 1)^.5 dA
Where Z = 2 - y

To get
2^.5 r dr d@
r is from 0 to R1
@ is from 0 to 2pi

Final answer is
(2)^.5*(pi)(R1)^2

Is this correct??

 

[LCKurtz]

Homework Statement

Cylinder : x^2 + y^2 = 1
Plane that intersects above cylinder: y + z = 2

What is the surface area of the sides of this cylinder?

Homework Equations

dS= R1 d@ dz
@ is from 0 to 2 pi
z is from 0 to 2 - y

dS=(Zx^2 + Zy^2 + 1)^.5 dA
Where Z = 2 - y

Please don't write Z when you mean z. Also your first relevant equation is $dS = R1 d\theta dz$. What is $R1$? Where does the equation I colored red come from and why do you need it since you already have $dS$? Anyway, you should set your integral up in terms of $\theta$ and $z$ as your first three lines suggest. And no, your answer is not correct.

 

[LaplacianHarmonic]

R1 is the constant radius of the cylinder. In this case it equals to the number 1.

Where does the equation I colored red come from and why do you need it since you already have $dS$?

The other equation for dS comes from the other derivation of dS in which I rotate a curve around the path of a circle in 3 dimensions.

Anyway, you should set your integral up in terms of $\theta$ and $z$ as your first three lines suggest.

Okay. I am just learning how to write correctly here. And no, your answer is not correct.

Can you help me? What did I do wrong?

 

[LCKurtz]

I thought I did help you. You have said:
dS= R1 d@ dz
@ is from 0 to 2 pi
z is from 0 to 2 - y
Where R1 = 1, so using proper notation you are saying:
$dS = d\theta dz$
$0\le \theta \le 2\pi$
$0 \le z \le 2-y$.
All you have to do is express $y$ in terms of $\theta$ and set up the double $dzd\theta$ integral you describe.

 

[LaplacianHarmonic]

I thought I did help you. You have said:
dS= R1 d@ dz
@ is from 0 to 2 pi
z is from 0 to 2 - y
Where R1 = 1, so using proper notation you are saying:
$dS = d\theta dz$
$0\le \theta \le 2\pi$
$0 \le z \le 2-y$.
All you have to do is express $y$ in terms of $\theta$ and set up the double $dzd\theta$ integral you describe.

So, 2 - sin@?

That gives me the value of 4pi as the final answer.

I already know there is something wrong with the answer. dS=d@dz

Does not come out from the

vector V= < 1cos@, 1sin@, z>
dS = <1cos@, 1sin@, 0> for the above vector eqn start.
dS=d@dz

Does not appear to have a vector equation...?

 

[LCKurtz]

Yes, that is correct. By the way, it is easy to type special characters like $\theta$. Just type ##\theta##.

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

Thread reopened after a post has been deleted.

 

[LCKurtz]

@LaplacianHarmonic : After reading your now deleted post, I am done trying to help you.

 

[berkeman]

And @LaplacianHarmonic -- please read the LaTeX tutorial in the Help/How-To section of the PF under INFO at the top of the page. It is very difficult trying to follow what you are writing when you try to type out equations as text. Thank you.

 

[berkeman]

And after a PM discussion with the OP, this thread will remain closed.