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Math Help - Finding the surface area of a cone?

  1. #1
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    Finding the surface area of a cone?

    Here is the question...

    Find the area of the surface of the cone z=1-r in the first octant.

    Can someone do this and explain to me how its done? I am unsure of where to start and what to do with the equation. Thanks in advance.
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  2. #2
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    Quote Originally Posted by Infernorage View Post
    Here is the question...

    Find the area of the surface of the cone z=1-r in the first octant.

    Can someone do this and explain to me how its done? I am unsure of where to start and what to do with the equation. Thanks in advance.
    Imagine dividing the first quadrant of the xy-plane, under the cone into small regions. The area of each region is \Delta A. Now imagine extending each region up to the cone. It has height z and so volume z\Delta A. The volume you want would be, approximately, the sum of those volumes over all the regions. That is, of course, a "Riemann sum" and, in the limit as the size of the regions goes to 0, you get the volume equal to the integral \int\int z dA.

    Since you have given the equation of the cone in terms of "r": z= 1- r rather than z^2= 1- x^2- y^2, I presume you want to use polar coordinates. To cover the entire base of the cone, obviously, r must go from 0 to 1 and to get the entire first quadrant, \theta must go from 0 to \pi/2. Don't forget that the "differential of area" in polar coordinates is dA= r drd\theta.

    The volume is given by \int_{r=0}^1\int_{\theta= 0}^{\pi/2} (1- r) r dr d\theta.
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  3. #3
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    Try doing it this way:

    You know the common formula for surface area right:

    S=\mathop\int\int\limits_{\hspace{-15pt}R} \sqrt{(f_x)^2+(f_y)^2+1}\; dydx

    and if the cone is given in polar coordinates z=1-r then the radius in x-y coordinates is just r=\sqrt{x^2+y^2}. Then the equation of the cone is c(x,y)=f(x,y)=1-\sqrt{x^2+y^2}. Plug that into the formula then and just do the part over the first quadrant to get the surface area in the first octant. The area R to integrate over is just a quarter circle right? You can do that.
    Last edited by shawsend; November 24th 2009 at 06:12 AM. Reason: only calculate area in first octant correction
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Imagine dividing the first quadrant of the xy-plane, under the cone into small regions. The area of each region is \Delta A. Now imagine extending each region up to the cone. It has height z and so volume z\Delta A. The volume you want would be, approximately, the sum of those volumes over all the regions. That is, of course, a "Riemann sum" and, in the limit as the size of the regions goes to 0, you get the volume equal to the integral \int\int z dA.

    Since you have given the equation of the cone in terms of "r": z= 1- r rather than z^2= 1- x^2- y^2, I presume you want to use polar coordinates. To cover the entire base of the cone, obviously, r must go from 0 to 1 and to get the entire first quadrant, \theta must go from 0 to \pi/2. Don't forget that the "differential of area" in polar coordinates is dA= r drd\theta.

    The volume is given by \int_{r=0}^1\int_{\theta= 0}^{\pi/2} (1- r) r dr d\theta.
    That all makes sense, but I am trying to find surface area, not volume. That integral wouldn't still apply, would it?
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  5. #5
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    Quote Originally Posted by Infernorage View Post
    That all makes sense, but I am trying to find surface area, not volume. That integral wouldn't still apply, would it?
    You are right. I completely misread your post!

    Shawsend is correct but in hopes of redeeming myself, here is a way to avoid having to change to xy-coordinates and set up that "complicated" square root. This does require a little more "mathematical sophistication".

    In polar coordinates, x= r cos(\theta) and y= r sin(\theta) so a "position vector" for any point on the surface of the cone can be written as \vec{v}(r,\theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ (1- r)\vec{k}.

    Two vectors tangent to the surface are \vec{v}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}- \vec{k} and \vec{v}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}. The cross product of those two vectors, -rcos(\theta)\vec{i}- rsin()\theta)\vec{j}+ r\vec{k}, sometimes called the "fundamental vector product" for the surface, is perpendicular to both and so normal to the surface and its length, \sqrt{r^2 cos^2(\theta)+ r^2 sin^2(\theta)+ r^2}= r\sqrt{2} gives the "differential of area": r\sqrt{2} drd\theta. Integrate that over the quarter circle to find the surface area.

    Here is a way of getting the surface area without integrating that I mentioned in another thread:

    Imagine cutting the cone along the line from the "tip" perpendicular to the circular base and "flattening" the cone out. (A cone is a "developable surface" and, unlike a hemisphere, can be flattened.) That will be a disk with a "wedge" taken out. Your cone has height 1 and base radius 1 and so slant height \sqrt{2}. Your flattened cone fits inside a circle of radius \sqrt{2} which has area \pi(\sqrt{2})^2= 2\pi. But that circle also has circumference 2\pi\sqrt{2} while the base of the cone is only [itex]2\pi[/itex]. That is, the flattened cone takes up only \frac{2\pi}{2\pi\sqrt{2}}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} of the circumference of the circle and so only \frac{\sqrt{2}}{2} of the area: the surface area of the cone is \frac{\sqrt{2}}{2}(2\pi)= \pi\sqrt{2}. (The area of the portion in the first octant is, of course, 1/4 of that.)

    Now, can you get that answer doing the integral as shawsend suggested and the way I suggested?
    Last edited by HallsofIvy; November 24th 2009 at 09:32 AM.
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