# Thread: surface area of a paraboloid

1. ## surface area of a paraboloid

Hi,

I tried to calculate the area of a paraboloid, say z=x^2+y^2, between z=0 and z=1. I thought that in order to do this calculation, I can think on the paraboloid as consisted of circles with radius sqrt(z), so the surface area would be given by the integral from 0 to 1 on 2*pi*sqrt(z) (the integrand is the perimeter of one circle).

I know for sure that this result is incorrect, because I have seen one exam in which the calculation is carried on by using surface integral of the first kind, which gives a different solution.

2. you can do the integration in your way but you need to use "arc length" instead of the z coordinate. That is, the integrand form is not 2*pi*sqrt(z)*dz, but 2*pi*sqrt(z)*ds, where ds is the arc element, you can find an expression of ds by z, but it is not a very simple form.

3. Originally Posted by fict
Hi,

I tried to calculate the area of a paraboloid, say z=x^2+y^2, between z=0 and z=1. I thought that in order to do this calculation, I can think on the paraboloid as consisted of circles with radius sqrt(z), so the surface area would be given by the integral from 0 to 1 on 2*pi*sqrt(z) (the integrand is the perimeter of one circle).

I know for sure that this result is incorrect, because I have seen one exam in which the calculation is carried on by using surface integral of the first kind, which gives a different solution.
Note how,

$1 = x^2 + y^2$

$0 = x^2 + y^2$

Which gives rise to $D \to x^2 + y^2 \le 1$

If I recall the equation for surface area is

$\iint_D \sqrt{ ( \frac{ \partial z }{ \partial x } )^2 + (\frac{ \partial z }{ \partial y } )^2 + 1 }$

In which case we get (by switching to polar co-ordinates)

$\int_0^{ 2 \pi } d \theta \int_0^1 r \sqrt{ 4r + 1 } dr$

4. Originally Posted by xxp9
you can do the integration in your way but you need to use "arc length" instead of the z coordinate. That is, the integrand form is not 2*pi*sqrt(z)*dz, but 2*pi*sqrt(z)*ds, where ds is the arc element, you can find an expression of ds by z, but it is not a very simple form.
You are correct, but I just wondered why my proposed solution is incorrect; it quite makes sense - building the paraboloid from many circles.

5. Originally Posted by fict
You are correct, but I just wondered why my proposed solution is incorrect; it quite makes sense - building the paraboloid from many circles.
Your integrand is not correct. You have simplified my double integral, but have not carried through the dS correctly.

6. Originally Posted by AllanCuz
Your integrand is not correct. You have simplified my double integral, but have not carried through the dS correctly.
Yes, I know that it is incorrect, I'm just wondering from the logical point of view why it is incorrect.

7. Originally Posted by fict
Yes, I know that it is incorrect, I'm just wondering from the logical point of view why it is incorrect.
Your idea is correct. It's just your integrand is not correct because it does not represent what you're saying it represents. You still need to project an element of area.

$\int_0^1 \sqrt{z} dz$ where $\sqrt{z} = \sqrt{x^2 + y^2 }$

This is a cone....Your integrating a cone from 0 to 1. What does this have to do with surface area? This will give you area under the cone.

8. AllanCuz is right. Your integrand results in the area of a cone. The problem is, your coordinate curves are not laid down well on the surface to give rise to areas when integrated. For example, the x,y coordinate curves on the xy plane is good, since you can integrate the form dxdy to get areas. In your case, your coordinate curves are circles and radial rays. They're orthogonal but z is not normalized, which means, if you draw the curves z=0, 0.1, 0.2, 0.3... on your surface, they're not evenly distributed.