Find the surface area of the cone z=2sqrt(x^2+y^2) and above a region in the xy-plane with area 5.
If anyone could help me with this problem, I would really appreciate it.
Thanks.
John Marcum
Hello, John,Originally Posted by OntarioStud
as you may know the term $\displaystyle \sqrt{x^2+y^2}$ describes the distance between the origin and a point in the xy-plane.
If you draw a plane cut through the cone along the z-axis it'll look like the diagram which I attached.
With your cone you get: $\displaystyle r = \sqrt{x^2+y^2}$.
The slope s is $\displaystyle s=\sqrt{r^2+4r^2}=r\cdot \sqrt{5}$.
The surface area of a cone contains the base circle and the area, which is produced by the sloping line:
$\displaystyle A_S=\pi r^2+\pi r s$
$\displaystyle A_{surface}=\pi r^2+ \pi \cdot r \cdot s =$$\displaystyle \pi(x^2+y^2)+\pi\sqrt{x^2+y^2}\cdot \sqrt{5(x^2+y^2}$
Notice: I've corrected my mistake. I'm sorry if the previous result caused any irritation.
Simplify to:$\displaystyle (1+\sqrt{5}) \cdot \pi \cdot (x^2+y^2)=(1+\sqrt{5}) \cdot \pi \cdot r^2$
This surface has the value 5. So you get: $\displaystyle 5=(1+\sqrt{5}) \cdot \pi \cdot r^2$. Solve for r and re-substitute. You'll get the equation:
$\displaystyle \sqrt{x^2+y^2}=\sqrt{\frac{5(\sqrt{5}-1)}{4\cdot \pi}$
Greetings
EB