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.
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 $\displaystyle \int_{r=0}^1\int_{\theta= 0}^{\pi/2} (1- r) r dr d\theta$.
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.
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, $\displaystyle x= r cos(\theta)$ 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 [itex]2\pi[/itex]. 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?