# VOlume and parabolic cylinder

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• July 2nd 2013, 06:44 AM
n22
VOlume and parabolic cylinder
Hi ,
I would very much appreciate help and guidance withi this .
It is quite tough.thanks.
Find the volume of the region bounded by the parabolic cylinder z=1-x^2 and the planes z=0,y=0 and y+z=2

(also I dont quite fully understand ..but do I always find the projections -why and how many projections do I need to find ?-sorry abt the stupid question-ineed to check)
• July 2nd 2013, 06:54 AM
Prove It
Re: VOlume and parabolic cylinder
Start by doing a sketch of your region.
• July 2nd 2013, 04:42 PM
n22
Re: VOlume and parabolic cylinder
Attachment 28714

Please see my sketch and add any neccessary changes ..i did this on paint
thanks
• July 2nd 2013, 05:58 PM
Prove It
Re: VOlume and parabolic cylinder
OK, that's a good start. Now since your region opens up along the y-axis, that would be a good first variable to determine bounds for. What bounds can you come up with for y?
• July 3rd 2013, 01:14 AM
n22
Re: VOlume and parabolic cylinder
Quote:

Originally Posted by Prove It
OK, that's a good start. Now since your region opens up along the y-axis, that would be a good first variable to determine bounds for. What bounds can you come up with for y?

would the limits for z be 0≤z≤2-y

and y goes from where to where? hintsplz
• July 3rd 2013, 03:45 AM
Prove It
Re: VOlume and parabolic cylinder
No, start with y, the bounds will depend on x and z.
Then work on z, the bounds will depend on x.
Then your x bounds will only be numbers.
• July 3rd 2013, 08:02 PM
n22
Re: VOlume and parabolic cylinder
Quote:

Originally Posted by Prove It
No, start with y, the bounds will depend on x and z.
Then work on z, the bounds will depend on x.
Then your x bounds will only be numbers.

I tried to make a projection onto the y-z plane to make it easier for visualization..please checkAttachment 28725
y limits: 0≤y≤(1+x²) got this from intersection of the planes..
z limits:0≤z≤(1-x²)
x limits:1≤x≤0
Please check.thanks
• July 3rd 2013, 08:26 PM
Prove It
Re: VOlume and parabolic cylinder
Quote:

Originally Posted by n22
I tried to make a projection onto the y-z plane to make it easier for visualization..please checkAttachment 28725
y limits: 0≤y≤(1+x²) got this from intersection of the planes..

No. Picture drawing lines parallel to the y-axis. They will be bounded by the planes y = 0 and y + z = 2. How can you write the second plane as y = something?

Quote:

z limits:0≤z≤(1-x²)
Correct.

Quote:

x limits:1≤x≤0
Since when is $\displaystyle 1 \leq 0$. Assuming you meant $\displaystyle 0 \leq x \leq 1$ then you would be correct.
• July 3rd 2013, 11:37 PM
n22
Re: VOlume and parabolic cylinder
[QUOTE=Prove It;791795]No. Picture drawing lines parallel to the y-axis. They will be bounded by the planes y = 0 and y + z = 2. How can you write the second plane as y = something?

would it be 0≤y≤2-z
how did you get x=0?? I seem to be getting x=√-1 which is obviously wrong.
when y=0,z=2
when y=2-z,z=0 these were then subbed into x=sqrt(1-z)

sorry about the stupid questions.
• July 3rd 2013, 11:44 PM
Prove It
Re: VOlume and parabolic cylinder
Yes $\displaystyle 0 \leq y \leq 2 - z$ is correct.

I got x = 0 because you're told the region is bounded by the plane x = 0. I was pointing out that while the bounds you got for x were wrong, you wrote them down the wrong way, since $\displaystyle 1 \leq x \leq 0$ means you are saying $\displaystyle 1 \leq 0$, which it obviously isn't.

Anyway, now that you have the correct bounds, set up your triple integral.