How to draw the following volume (not using matlab)

Thread Starter

u-will-neva-no

Joined Mar 22, 2011
230
Hey again!

I have two surfaces: \(z = \sqrt{2}(x^2+y^2) (1)\) and \( x^2 + y^2 + z^2 = 1 (2)\)

now I know that (1) is the shape of a paraboloid and (2) is a sphere but how do the two look together?

I put \(x =0\) and get \(z = \sqrt{2}\) and \(z = \frac{-1}{\sqrt{2}}\) but for \(y=0\) I get an imaginary value for the x value (and also 4 solutions which is strange)
 

MrChips

Joined Oct 2, 2009
23,081
You know the equation of a circle is
\(x^2 + y^2 = r^2\)

Thus,
\(x^2 + y^2 = 1\)
is the equation of a circle with radius = 1


Similarly,
\(x^2 + y^2 + z^2= 1\)
is the equation of a sphere with radius = 1

See if you can figure out the shape of
\(x^2 + y^2 = z\)
 

panic mode

Joined Oct 10, 2011
1,877
1) z=sqrt(2)(x^2+y^2)
2) x^2+y^2+z^2=1


rewrite last one as
x^2+y^2=1-z^2
and substitute in first one to get quadratic equation
z=sqrt(2)(1-z^2)

with solutions z=1/sqrt(2) and z=-sqrt(2)
which are z values where surfaces intersect.

use that z-value in both equations to get curves of intersection (circles).

then freeze any one of variables to get section view.

with few points calculated, just connect the dots.
 

MrChips

Joined Oct 2, 2009
23,081
just put square braces around tex
in front of the equation
and close with square braces around /tex

like this

{tex}z=sqrt2(1-z^2){/tex}

but replace { } with [ ]

and you will get

\(z=sqrt2(1-z^2)\)
 
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