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

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- Jul 30th 2006, 07:11 PMOntarioStudSurface 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 - Jul 31st 2006, 04:50 AMearbothQuote:

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 - Jul 31st 2006, 10:51 AMThePerfectHacker
earboth which software you use for functions of two variables?

- Jul 31st 2006, 09:19 PMearbothQuote:

Originally Posted by**ThePerfectHacker**

Derive 6

Greetings

EB