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)

Re: VOlume and parabolic cylinder

Start by doing a sketch of your region.

1 Attachment(s)

Re: VOlume and parabolic cylinder

Attachment 28714

Please see my sketch and add any neccessary changes ..i did this on paint

thanks

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?

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

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.

1 Attachment(s)

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

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 check

Attachment 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.

Since when is $\displaystyle \displaystyle 1 \leq 0$. Assuming you meant $\displaystyle \displaystyle 0 \leq x \leq 1$ then you would be correct.

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.

Re: VOlume and parabolic cylinder

Yes $\displaystyle \displaystyle 0 \leq y \leq 2 - z$ 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 $\displaystyle \displaystyle 1 \leq x \leq 0$ means you are saying $\displaystyle \displaystyle 1 \leq 0$, which it obviously isn't.

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