# Area integral with cylindrical coordinates

1. May 13, 2013

### Nikitin

1. The problem statement, all variables and given/known data
find the area of the surface defined by x2+y2=y, with yE[0,4]

3. The attempt at a solution

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

402pi0r*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?

2. May 13, 2013

### LCKurtz

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

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?

3. May 13, 2013

### Staff: Mentor

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

4. May 13, 2013

### Staff: Mentor

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.

5. May 14, 2013

### Nikitin

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

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

6. May 14, 2013

### 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 $r d\theta dy$.

7. May 14, 2013

### 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?

8. May 14, 2013

### LCKurtz

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.

9. May 15, 2013

### Nikitin

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