# another triple integration!

• Aug 24th 2007, 07:41 PM
kittycat
another triple integration!
Use a triple integral to find the volume of the solid bounded by the surface y=x^2 and the planes y+z=4 and z=0.

Thank you very much.
• Aug 25th 2007, 02:56 AM
galactus
Try this:

$2\int_{0}^{2}\int_{x^{2}}^{4}\int_{0}^{4-y}dzdydx$

Here's a haphazard attempt at a graph of your region. I hope it gives you some idea of what it looks like.
• Aug 25th 2007, 09:32 AM
kittycat
Hi galactus,

Thank you very much.

Why is the integration of dy from x^2 to 4 ?
• Aug 25th 2007, 10:05 AM
galactus
Hello kittykat:

Because that's where the plane 'slices' the parabola x^2.

Here's a graph. The line y=4 is where the plane z=4-y meets the parabola.

So, you integrate over y from x^2 to 4. See?.

At x=2, y=x^2=4

The 3-D I posted is rather cock-eyed, but you can see it there.:)

I gonna go mow grass now, so I may not be back for a few hours.:(
• Aug 25th 2007, 10:36 AM
kittycat
" I gonna go mow grass now, so I may not be back for a few hours.:( "

;) Enjoy the green and beautiful grass!
• Aug 25th 2007, 10:39 AM
galactus
Actually, it's rather hot here in PA today.

Did you follow what I attempted to explain?.
• Aug 25th 2007, 10:52 AM
kittycat
Not really, I am still thinking ....

I am very bad in visualize the 3D objects. Thus , it is so how difficult for me to set up the integrals for triple integration. :(

Thanks! I got it now!
• Aug 25th 2007, 11:01 AM
galactus
First, try drawing your parabola y=x^2. That's easy enough. It's the plane rising up the z-axis that's the booger to visualize. Try plotting various points for it.

z=4-y, if y=0, z=4

If y=4, z=0. See there?. That's where the plane intersects the parabola in the xy plane.

Then it rises up the z-axis at a slant and intersects the z axis at y=0.