1. Surface Area Question

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

2. Originally Posted by OntarioStud
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,

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

3. earboth which software you use for functions of two variables?

4. Originally Posted by ThePerfectHacker
earboth which software you use for functions of two variables?
Hello, TPH,

Derive 6

Greetings

EB