Surface of the cone inside the cylinder

**Metrika** · November 26th 2010, 04:56 PM

Help me compose the surface integral to solve this problem, please

Calculate the area of the surface of the cone $x^2-y^2-z^2=0$ , located inside the cylinder $x^2+y^2=1$ .
Make a drawing.

**DeMath** · November 26th 2010, 05:22 PM

Originally Posted by Metrika

Help me compose the surface integral to solve this problem, please

Calculate the area of the surface of the cone $x^2-y^2-z^2=0$ , located inside the cylinder $x^2+y^2=1$ .
Make a drawing.

See drawing; but I do not know how to compose the integral to your problem((

**Metrika** · November 26th 2010, 08:21 PM

omg! stunningly beautiful!
In what program this is done?

Is WolframAlpha?

**HallsofIvy** · November 27th 2010, 03:39 AM

So that we can write this in regular polar cylindrical coordinates, I am going to swap "x" and "z" in the problem:
$z^2- x^2- y^2= 0$ and $z^2+ y^2= 1$ .

In cylindrical coordinates, that is $z^2= r^2$ and $z^2+ r^2 sin^2(\theta)= 1$ so the two surfaces intersect where $r^2+ r^2 sin^2(\theta)= 1$ or $r= \pm \sqrt{\frac{1}{1+ sin^2(\theta)}$ . Integrate with $\theta$ from 0 to $2\pi$ and, for each $\theta$ , r from 0 to $\frac{1}{\sqrt{1+ sin^2(\theta)}}$ .

The "differential of surface area" for the cone is $\sqrt{2} r drd\theta$ .

**DeMath** · November 27th 2010, 04:21 AM

Originally Posted by Metrika

omg! stunningly beautiful!
In what program this is done?

Is WolframAlpha?

I did it with the Maple_13; нere's the code for building

A := plot3d([[r*cos(t),r*sin(t),r*sqrt(cos(2*t))],[r*cos(t),r*sin(t),-r*sqrt(cos(2*t))]], t=-(1/4)*Pi .. (1/4)*Pi, r=0 .. 1, numpoints=3000, color="LightBlue", style=surface):
B := plottools[rotate](A,0,0,Pi):
C := plot3d([cos(t),sin(t),z], t=0 .. 2*Pi, z=-1 .. 1, color=pink, style=wireframe, transparency=0.75, numpoints=2000):
plots[display](A,B,C, axes=normal, scaling=constrained, lightmodel=light2, view=[-1.4 .. 1.4, -1.4 .. 1.4, -1.4 .. 1.4], orientation=[70,62]);

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