Area integral with cylindrical coordinates

[Nikitin]

Homework Statement

find the area of the surface defined by x²+y²=y, with y_E[0,4]

The Attempt at a Solution

I tried setting it up with cylindrical coordinates, but it doesn't work. Why?

∫⁴₀∫^2pi₀r*dθ*dy, where r=√y

Is it because my height, dy, has a vertical direction while its direction should be in the direction of the conic wall?

 

[LCKurtz]

Homework Statement

find the area of the surface defined by x²+y²=y, with y_E[0,4]

What does $y_E[0,4]$ mean?

The Attempt at a Solution

I tried setting it up with cylindrical coordinates, but it doesn't work. Why?

∫⁴₀∫^2pi₀r*dθ*dy, where r=√y

Is it because my height, dy, has a vertical direction while its direction should be in the direction of the conic wall?

For one thing, $y$ is not one of the cylindrical coordinates. Your first step should be to draw a picture of your surface. Have you done that? What do you get?

 

[Mark44]

What does $y_E[0,4]$ mean?

I'm going to guess that it means $y \in [0, 4]$.

 

[Mark44]

Homework Statement

find the area of the surface defined by x²+y²=y, with y_E[0,4]

Are you sure you have copied the problem correctly? What you have doesn't make sense to me. In R², the equation above represents a circle. In R³, the equation represents a circular cylinder that extends infinitely far on both directions along the z-axis.

Did you mean $z \in [0, 4]$? That would provide upper and lower bounds for the cylinder, so you could get a finite surface area.

The Attempt at a Solution

I tried setting it up with cylindrical coordinates, but it doesn't work. Why?

∫⁴₀∫^2pi₀r*dθ*dy, where r=√y

Is it because my height, dy, has a vertical direction while its direction should be in the direction of the conic wall?

 

[Nikitin]

crap, I meant x^2 +z^2 = y..

Sorry, that was really stupid of me. It's in |R^3

 

[LCKurtz]

OK, then cylindrical coordinates with the variables changed makes sense. But you still need to get the correct formula for $dS$. If you parameterize your surface as $\vec R(\theta,y)$ then you need to use the formula $dS = |\vec R_\theta \times \vec R_y|d\theta dy$. It isn't $r d\theta dy$.

 

[Nikitin]

Yeah, I ultimately did that to solve my problem. However, I don't understand why the cylindrical formula shouldn't work.. If you just add the areas of lots of really thin cylinders together, you should end up with the area of the cone, right?

 

[LCKurtz]

Yeah, I ultimately did that to solve my problem. However, I don't understand why the cylindrical formula shouldn't work.. If you just add the areas of lots of really thin cylinders together, you should end up with the area of the cone, right?

You lost me there. For one thing, the surface isn't a cone. I don't know what thin cylinders you are talking about. You need to add up elements of surface area.

 

[Nikitin]

ah, you're right, it isn't a cone. I'm sorry, never mind this thread.