final tomorrow...triple integral help please!

Dubulus · #1 $Old$ August 6th, 2008, 02:33 PM

i need to find the volume between the paraboloid z= 4x^2 + 4y^2 and the plane z=4. im having troubles deciding which coordinates i should use, and what the boundaries would be.

galactus · #2 $Old$ August 6th, 2008, 02:48 PM

The plane at z=4 projects a circle of radius 1 onto the xy-plane.

Using polar coordinates, $z=4r^{2}$

$\int_{0}^{2\pi}\int_{0}^{1}\int_{4r^{2}}^{4}rdzdrd{\theta}$

Chris L T521 · #3 $Old$ August 6th, 2008, 03:44 PM

Quote:

Originally Posted by Dubulus $View Post$

i need to find the volume between the paraboloid z= 4x^2 + 4y^2 and the plane z=4. im having troubles deciding which coordinates i should use, and what the boundaries would be.

Quote:

Originally Posted by galactus $View Post$

The plane at z=4 projects a circle of radius 2 onto the xy-plane.

Using polar coordinates, $z=4r^{2}$

$\int_{0}^{2\pi}\int_{0}^{2}\int_{4}^{4r^{2}}rdzdrd{\theta}$

Um...

at $z=4$ , the cross sectional region would be $4x^2+4y^2=4\implies x^2+y^2=1$ which would imply a circle of radius 1.

I agree with everything else that you have Galactus, except the limits for r.

So my integral setup would be $\int_{0}^{2\pi}\int_{0}^{{\color{red}1}}\int_{4}^{4r^{2}}r \,dz\,dr\,d{\theta}$

--Chris

galactus · #4 $Old$ August 6th, 2008, 04:14 PM

Yes, you are correct Chris. Brain fart/typo. That's what I get for doing a triple integral and watching TV at the same time.

$4r^{2}=4\Rightarrow r=1$

JaneBennet · #5 $Old$ August 6th, 2008, 04:15 PM

The cross sections with the planes parallel to $z=4$ are all circles. Therefore you can find the volume just by rotating the curve $z=4y^2$ about the z-axis through $2\pi$ radians. ( $z=4y^2$ is the equation of the cross section of the paraboloid with the yz-plane.)

So the volume is $\pi\int_0^4{y^2}\,dz=\pi\int_0^4{\frac{z}{4}}\,dz=\pi\left[\frac{z^2}{8}\right]_0^4=2\pi$ .

There. No need for triple integration.

galactus · #6 $Old$ August 6th, 2008, 04:42 PM

Jane, I reckon Chris and I are in error together. I get 2Pi

Here it is in rectangular and I get the same result.

$\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{4x^{2}+4y^{2}}^{4}dzdydx=2{\pi}$

JaneBennet · #7 $Old$ August 6th, 2008, 04:52 PM

You’re right, the answer is $2\pi$ . I mistyped the formula for volume of solid of revolution as $2\pi\int_a^b{y^2}\,dx$ rather than $\pi\int_a^b{y^2}\,dx$ .

But my point is that you don’t need to use triple integration for this problem. You just need to use some lateral thinking.

galactus · #8 $Old$ August 6th, 2008, 04:56 PM

Yes, very good. I was going on the assumption the poster had to use triple

integration. Perhaps because that is what they are studying now. Even so, if

a student had the foresight and lateral thinking to answer a problem in that

method, surely the instructor would give them credit and a kudo.