# Thread: Finding the surface area of a cone?

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

Here is the question...

Find the area of the surface of the cone $\displaystyle z=1-r$ 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.

2. Originally Posted by Infernorage
Here is the question...

Find the area of the surface of the cone $\displaystyle z=1-r$ 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 $\displaystyle \Delta A$. Now imagine extending each region up to the cone. It has height z and so volume $\displaystyle z\Delta A$. 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 $\displaystyle \int\int z dA$.

Since you have given the equation of the cone in terms of "r": z= 1- r rather than $\displaystyle z^2= 1- x^2- y^2$, 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, $\displaystyle \theta$ must go from 0 to $\displaystyle \pi/2$. Don't forget that the "differential of area" in polar coordinates is $\displaystyle dA= r drd\theta$.

The volume is given by $\displaystyle \int_{r=0}^1\int_{\theta= 0}^{\pi/2} (1- r) r dr d\theta$.

3. Try doing it this way:

You know the common formula for surface area right:

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

and if the cone is given in polar coordinates $\displaystyle z=1-r$ then the radius in x-y coordinates is just $\displaystyle r=\sqrt{x^2+y^2}$. Then the equation of the cone is $\displaystyle c(x,y)=f(x,y)=1-\sqrt{x^2+y^2}$. 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.

4. Originally Posted by HallsofIvy
Imagine dividing the first quadrant of the xy-plane, under the cone into small regions. The area of each region is $\displaystyle \Delta A$. Now imagine extending each region up to the cone. It has height z and so volume $\displaystyle z\Delta A$. 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 $\displaystyle \int\int z dA$.

Since you have given the equation of the cone in terms of "r": z= 1- r rather than $\displaystyle z^2= 1- x^2- y^2$, 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, $\displaystyle \theta$ must go from 0 to $\displaystyle \pi/2$. Don't forget that the "differential of area" in polar coordinates is $\displaystyle dA= r drd\theta$.

The volume is given by $\displaystyle \int_{r=0}^1\int_{\theta= 0}^{\pi/2} (1- r) r dr d\theta$.
That all makes sense, but I am trying to find surface area, not volume. That integral wouldn't still apply, would it?

5. Originally Posted by Infernorage
That all makes sense, but I am trying to find surface area, not volume. That integral wouldn't still apply, would it?

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

Two vectors tangent to the surface are $\displaystyle \vec{v}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}- \vec{k}$ and $\displaystyle \vec{v}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}$. The cross product of those two vectors, $\displaystyle -rcos(\theta)\vec{i}- rsin()\theta)\vec{j}+ r\vec{k}$, sometimes called the "fundamental vector product" for the surface, is perpendicular to both and so normal to the surface and its length, $\displaystyle \sqrt{r^2 cos^2(\theta)+ r^2 sin^2(\theta)+ r^2}= r\sqrt{2}$ gives the "differential of area": $\displaystyle r\sqrt{2} drd\theta$. 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 $\displaystyle \sqrt{2}$. Your flattened cone fits inside a circle of radius $\displaystyle \sqrt{2}$ which has area $\displaystyle \pi(\sqrt{2})^2= 2\pi$. But that circle also has circumference $\displaystyle 2\pi\sqrt{2}$ while the base of the cone is only $2\pi$. That is, the flattened cone takes up only $\displaystyle \frac{2\pi}{2\pi\sqrt{2}}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$ of the circumference of the circle and so only $\displaystyle \frac{\sqrt{2}}{2}$ of the area: the surface area of the cone is $\displaystyle \frac{\sqrt{2}}{2}(2\pi)= \pi\sqrt{2}$. (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?