1. ## Volume Integral

Question:
The volume V between
$\displaystyle z = \frac{5{x}^{2} + 3{y}^{2}}{R}$

And the x, y plane (z = xy?)

What is the volume of this region?

Thoughts?
$\displaystyle \iiint dV = \iiint dx dy dz$
Given that this is a paraboloid (3D) im not too sure how to approach it. I guess it just wants the volume of the paraboloid not entirely sure if the z = xy changes anything seeming the paraboloid doesn't go into negative z anyway?

Any help on how i should approach and deal with this is much appreciated!

2. ## Re: Volume Integral

Originally Posted by imagemania
Question:
The volume V between
$\displaystyle z = \frac{5{x}^{2} + 3{y}^{2}}{R}$

And the x, y plane (z = xy?)

What is the volume of this region?

Thoughts?
$\displaystyle \iiint dV = \iiint dx dy dz$
Given that this is a paraboloid (3D) im not too sure how to approach it. I guess it just wants the volume of the paraboloid not entirely sure if the z = xy changes anything seeming the paraboloid doesn't go into negative z anyway?

Any help on how i should approach and deal with this is much appreciated!
What is R?

3. ## Re: Volume Integral

Maybe $\displaystyle z=x+y$ or $\displaystyle z=x-y$??

4. ## Re: Volume Integral

Sorry, R is a constant

5. ## Re: Volume Integral

The xy plane is the plane given by z = 0 .

6. ## Re: Volume Integral

Sorry im still not following,

I understand teh diagram looks of the order of this:
http://img202.imageshack.us/img202/4418/captureidmd.png

And this questions asks for the volume between that and the xy plane i.e. the infinite square that supports it.

Though i can't see converting the coordinate system as useful. I thought about doing:
$\displaystyle \vec{N} = r_{x} \times r_{y}$
Where:
$\displaystyle r_{x} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial x} ...$
but then N is a vector and not useful here as i want scalars!

7. ## Re: Volume Integral

Originally Posted by imagemania
Question:
The volume V between
$\displaystyle z = \frac{5{x}^{2} + 3{y}^{2}}{R}$

And the x, y plane

What is the volume of this region?
Any help on how i should approach and deal with this is much appreciated!
Maybe the author of this problem implied such equations of planes: $\displaystyle z=x,~z=y$??

Then we can calculate the finite volume.

8. ## Re: Volume Integral

I'll word the question fully, V will be the volume
Consider V inside the cylinder
$\displaystyle x^{2} + y^{2} = 6{R}^{2}$
and between $\displaystyle z = (\frac{5{x}^{2} + 3{y}^{2}}{R})$ and the (x, y) plane.

x, y and z are the Cartesian coordinates and R is also a constant.

Now i read this question and thought it meant "Find the volume of the cylinder then find teh volume betwen the other two independently" i.e. two separate questions. Perhaps i was wrong?

Finding V for teh cylinder is just a matter of finding the jacobian and using cylindrical coordinates surely?

But the other part, assuming it is a separate question, i am not sure how to best approach it

Thanks

9. ## Re: Volume Integral

The (x, y) plane given by z=0. So

\usepackage[usenames]{color}\gammacorrection{1.5}\usepackage[11pt]{extsizes}{\begin{aligned}T&=\left\{(x,y,z)\in\mathbb{R}^3\mid\,x^2+y^2\leqslant6R^2,~0\leqslant z\leqslant\frac{5x^2+3y^2}{R}\right\}\\[7pt] V&=\iiint\limits_T{dxdydz}=\iint\limits_{x^2+y^2\leqslant6R^2}dxdy\int\limits_0^{\tfrac{5x^2+3y^2}{R}}dz=\frac{1}{R}\iint\limits_{x^2+y^2\leqslant6R^2}(5x^2+3y^2)\,dxdy=\\ &=\left\{\begin{gathered}x=r\cos\theta,\hfill\\y=r\sin\theta\hfill\end{gathered}\right\}=\frac{1}{R}\int\limits_0^{2\pi}\int\limits_0^{\sqrt6R}(5r^2\cos^2\theta+3r^2\sin^2\theta)r\,drd\theta=\\ &=\frac{1}{R}\int\limits_0^{2\pi}(5\cos^2\theta+3\sin^2\theta)\,d\theta\int\limits_0^{\sqrt6R}r^3\,dr=\frac{1}{R}\int\limits_0^{2\pi}(2\cos^2\theta+3)\,d\theta\cdot\left.{\frac{r^4}{4}}\right|_0^{\sqrt6R}=\\ &=\frac{(\sqrt6R)^4}{4R}\int\limits_0^{2\pi}(\cos2\theta+4)\,d\theta=\left.{9R^3\left(\frac{1}{2}\sin2\theta+4\theta\right)}\!\right|_0^{2\pi}=9R^3\cdot8\pi=72\pi R^3\end{aligned}}

See the picture

10. ## Re: Volume Integral

Oh i see now DeMath, it is simply using the first as z the and then introducing the cylindrical coordinates - didn't help i miss understood the question thanks a lot

Though i must find out, how did you draw that image? I currently use Maple (14) and haven't figured how to draw them within the same body, what did you use and how did you do it?

11. ## Re: Volume Integral

Originally Posted by imagemania
Though i must find out, how did you draw that image? I currently use Maple (14) and haven't figured how to draw them within the same body, what did you use and how did you do it?
You're lucky - this is done in Maple
Here is the code of this body (for $\displaystyle R=\sqrt{6}$)

plot3d([(5x^2+3y^2)/sqrt(6)], x=-6..6, y=-sqrt(36-x^2)..sqrt(36-x^2), filled=true, style=hidden, color="Cyan", lightmodel=light2, transparency=0.25, numpoints=10000, axes=normal, orientation=[81, 60])

When you see the picture, then move the cursor on her, click the right key of the mouse and after Esc.