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

Discussion in 'Math' started by u-will-neva-no, Mar 28, 2012.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
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)

2. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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PM 1Chance...She can help.

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3. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Sorry, I don't understand what you wrote.

4. ### R!f@@ AAC Fanatic!

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

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5. ### MrChips Moderator

Oct 2, 2009
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3,362
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
$x^2 + y^2 = z$

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6. ### panic mode Senior Member

Oct 10, 2011
1,320
304
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.

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7. ### MrChips Moderator

Oct 2, 2009
12,446
3,362
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

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

8. ### panic mode Senior Member

Oct 10, 2011
1,320
304
test:

$sqrt(1-z^2)$

$sqrt x^5$

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

cool ;-)

thanks

9. ### Georacer Moderator

Nov 25, 2009
5,142
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