Volume between 2 paraboloids

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June 19th 2009, 06:37 PM
arbolis

Volume between 2 paraboloids

Today was my test and I've absolutely no idea about if I'll pass it or not. (On the cut, or I don't know how to say)
I had to calculate the volume between the paraboloids $z=x^2+y^2$ and $z=\frac{4}{3}-\frac{x^2}{3}-\frac{y^2}{3}$ , if I remember well.
I've thought about it and I think that using Cartesian coordinates will be easier than spherical and cylindrical ones.
However I tried to calculate the volume via cylindrical coordinates: $(r,\theta,z)$ . I tried to set up the triple integral : $z$ goes from $r^2$ to $\frac{4}{3}-\frac{r^2}{3}$ .
$\theta$ from $0$ to $2 \pi$ .
And $r$ goes from $0$ to .... I wasn't sure. But I strongly believe I already made an error.

I have 2 questions : How can I calculate the volume they ask using cylindrical coordinates?
With what method would you solve the problem?
Thank you very much!

Oh, and before I forget, I know they cross each other with a circular ( $x^2+y^2=1$ ) section.
June 19th 2009, 06:55 PM
Amer

Quote:

Originally Posted by arbolis

Today was my test and I've absolutely no idea about if I'll pass it or not. (On the cut, or I don't know how to say)
I had to calculate the volume between the paraboloids $z=x^2+y^2$ and $z=\frac{4}{3}-\frac{x^2}{3}-\frac{y^2}{3}$ , if I remember well.
I've thought about it and I think that using Cartesian coordinates will be easier than spherical and cylindrical ones.
However I tried to calculate the volume via cylindrical coordinates: $(r,\theta,z)$ . I tried to set up the triple integral : $z$ goes from $r^2$ to $\frac{4}{3}-\frac{r^2}{3}$ .
$\theta$ from $0$ to $2 \pi$ .
And $r$ goes from $0$ to .... I wasn't sure. But I strongly believe I already made an error.

I have 2 questions : How can I calculate the volume they ask using cylindrical coordinates?
With what method would you solve the problem?
Thank you very much!

Oh, and before I forget, I know they cross each other with a circular ( $x^2+y^2=1$ ) section.

in Cartesian coordinate

$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int^{\frac{4-x^2-y^2}{3}}_{x^2+y^2} dzdydx$

in polar

$\int_{0}^{2\pi}\int_{0}^{1}\int_{r^2}^{\frac{4-r^2}{3}}r dzdrd\theta$

Edit : I forget the jacobian
is there anything is not clear ??
June 19th 2009, 07:29 PM
Amer

2 Attachment(s)

to determine the boundaries I prefer to sketch the graph like this

Attachment 11944

and I put in it the important points then I see z change from which curve to which curve in our graph we see that z change from $x^2+y^2 ,to...,\frac{4-x^2-y^2}{3}$
since the graph of the first curve is lower than the second one after that we do a dropping to the x-y plane since we first determine the boundaries of z if we drop of the whole solid that we have we will get a circle with radius 1 this is the intersect between two paraboloids like this

Attachment 11945

from the picture we see that y change from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$

then x change from -1 to 1

in polar it is the same for z and for r and theta I think it is clear

I am sorry for my bad graph
June 19th 2009, 08:15 PM
arbolis

Oh... I wasn't far at all! I think I even tried it, but I made an arithmetic mistake and I gave up on this exercise.
Thanks Amer, everything is clear.