Help me compose the surface integral to solve this problem, please
Calculate the area of the surface of the cone $\displaystyle x^2-y^2-z^2=0$, located inside the cylinder $\displaystyle x^2+y^2=1$.
Make a drawing.
Help me compose the surface integral to solve this problem, please
Calculate the area of the surface of the cone $\displaystyle x^2-y^2-z^2=0$, located inside the cylinder $\displaystyle x^2+y^2=1$.
Make a drawing.
So that we can write this in regular polar cylindrical coordinates, I am going to swap "x" and "z" in the problem:
$\displaystyle z^2- x^2- y^2= 0$ and $\displaystyle z^2+ y^2= 1$.
In cylindrical coordinates, that is $\displaystyle z^2= r^2$ and $\displaystyle z^2+ r^2 sin^2(\theta)= 1$ so the two surfaces intersect where $\displaystyle r^2+ r^2 sin^2(\theta)= 1$ or $\displaystyle r= \pm \sqrt{\frac{1}{1+ sin^2(\theta)}$. Integrate with $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$ and, for each $\displaystyle \theta$, r from 0 to $\displaystyle \frac{1}{\sqrt{1+ sin^2(\theta)}}$.
The "differential of surface area" for the cone is $\displaystyle \sqrt{2} r drd\theta$.
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]);