# How to draw the following volume (not using matlab)

#### 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)

#### R!f@@

Joined Apr 2, 2009
9,747
PM 1Chance...She can help.

#### u-will-neva-no

Joined Mar 22, 2011
230
Sorry, I don't understand what you wrote.

#### R!f@@

Joined Apr 2, 2009
9,747
1Chance is the math wiz of AAC.

#### 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

$$z=sqrt2(1-z^2)$$

#### panic mode

Joined Oct 10, 2011
1,877
test:

$$sqrt(1-z^2)$$

$$sqrt x^5$$

$$c=sqrt (a^2+b^2)$$

cool ;-)

thanks