Surface Integral: Find Area of Cone (0 ≤ z ≤ 4)

[MacLaddy]

Homework Statement

Find the area of the following surface using an explicit description of a surface.

The cone z^2=4x^2+4y^2\mbox{ for } 0 \leq z \leq 4

Homework Equations

\iint_s f(x,y,z)dS=\iint_R f(x,y,g(x,y))\sqrt{z^2_x+z^2_y +1}

The Attempt at a Solution

I have solved the dS portion of this, and it is \sqrt{5}, however, I can not seem to figure out my limits of integration.

\iint \sqrt{5}dA should be my set-up?

From geometry I know the answer is 4\pi\sqrt{5}, but I can't seem to get it via integration.

The radius at the top is 2, the height is 4, but the radius is also variable as you go down the cone, so it needs a third parameter. However, this should only be done with a double integral, so I am lost.

Any help is appreciated.

Thanks,
Mac

P.S. Any advice on how to keep my Latex in line with my sentences would also be appreciated.

 

[Dick]

Homework Statement

Find the area of the following surface using an explicit description of a surface.

The cone z^2=4x^2+4y^2\mbox{ for } 0 \leq z \leq 4

Homework Equations

\iint_s f(x,y,z)dS=\iint_R f(x,y,g(x,y))\sqrt{z^2_x+z^2_y +1}

The Attempt at a Solution

I have solved the dS portion of this, and it is \sqrt{5}, however, I can not seem to figure out my limits of integration.

\iint \sqrt{5}dA should be my set-up?

From geometry I know the answer is 4\pi\sqrt{5}, but I can't seem to get it via integration.

The radius at the top is 2, the height is 4, but the radius is also variable as you go down the cone, so it needs a third parameter. However, this should only be done with a double integral, so I am lost.

Any help is appreciated.

Thanks,
Mac

P.S. Any advice on how to keep my Latex in line with my sentences would also be appreciated.

You just want to integrate dxdy over the region in the xy plane that your surface lies over, in this case the interior of the circle $x^2+y^2=4$, right? You can do it in xy coordinates, but it's easier in polar. To keep your tex on the same line use the 'itex' delimiter instead of 'tex'.

 

[MacLaddy]

You just want to integrate dxdy over the region in the xy plane that your surface lies over, in this case the interior of the circle $x^2+y^2=4$, right? You can do it in xy coordinates, but it's easier in polar. To keep your tex on the same line use the 'itex' delimiter instead of 'tex'.

Well then, that's unbelievably simple... \int_0^{2\pi}\int_0^2\sqrt{5}rdrd\theta=4\pi\sqrt{5}. It's almost like I forgot how to do polar for a minute. (thanks for the itex tip)

However, this confuses me. If all I'm integrating is the circle at the top, then how does \sqrt{5} give me the surface area? That integration gives the area of that top circle, not even the volume of the cone.

Sorry, it's not intuitive for me.

Thanks for the help,
Mac

 

[Dick]

Well then, that's unbelievably simple... \int_0^{2\pi}\int_0^2\sqrt{5}rdrd\theta=4\pi\sqrt{5}. It's almost like I forgot how to do polar for a minute. (thanks for the itex tip)

However, this confuses me. If all I'm integrating is the circle at the top, then how does \sqrt{5} give me the surface area? That integration gives the area of that top circle, not even the volume of the cone.

Sorry, it's not intuitive for me.

Thanks for the help,
Mac

You aren't really integrating over the 'circle at the top'. Everything in your surface integral is a function of x and y. So you want to integrate over all x and y such that (x,y,z(x,y)) with z(x,y)=sqrt(4*x^2+4*y^2) is on the part of the cone you are considering. It's just that circle of radius 2 in the xy plane.

 

[MacLaddy]

Interesting. I think I'll have to run through the proof once or twice to understand fully.

Thanks again.
Mac