Integration of multiple variables

ElijahRockers

1. The problem statement, all variables and given/known data

Find the volume of the given solid.
Under the plane x − 2y + z = 8 and above the region bounded by x + y = 1 and x² + y = 1

3. The attempt at a solution

Here's how I set it up.

[itex]\int^1_0 \int^{1-x^2}_{1-x} (8-x+2y) dydx[/itex]

When I do the math, I get 21/20. I have gone several different routes using a calculator and I keep getting that answer. The software tells me the answer is 29/20.

So am I setting it up wrong?

Dick

For what it's worth, I get 21/20 as well.

ElijahRockers

Well, I copied and pasted the question directly from the software this time. Stupid software.

Dick

Well, assuming we are right, I doubt the software figured out the answer. Somebody told it what the answer should be. They are to blame.

ElijahRockers

Hmm I found an error in my work. The anti derivative of 8-x+2y is 8x-xy+y^2.

I had 8x-xy-y^2.

Dick

I think it's 8y-xy-y^2. Are you changing your answer? I'm not.

ElijahRockers

But if x-2y+z=8, then z=8-x+2y dz/dy = 8y -xy +y^2. I haven't gone through the problem, I did a similar one with different numbers and got it right

Dick

You are integrating z, not differentiating it. I thought you did go through it and got 21/20?? I'm not sure where you are going with this. I think you did it right the first time.

ElijahRockers

Oh yeah, sorry, by dz/dy i mean the anti derivative.

And yes, i got 21/20 when i used 8x-xy-y^2. I noticed my error, but by that time had already gotten the question right with a different set of values.

I posted my error here just incase you were wondering, and to see if, by some bizarre one in a million coincidence, you made the same exact error I did. (not likely, especially considering you are probably much more careful than I am)

Dick

Wow. Apparently I'm not much more careful than you. I made a completely different error. I integrated z=8-x-2y instead of z=8-x+2y and got 21/20. Figured since I got the same as you did, no need to double check. Oooops. Yeah, the odds are pretty low for this to happen.

ElijahRockers

........Actually, the odds drop drastically when we consider that you actually didn't make a completely different error, but you made the same exact error I did, completely separately from me. Look at the post where I mentioned my error.

Dick

Ok, yeah. I get you, it's wasn't the 8x instead of the 8y, it was the sign on the 2y. That does cut the odds.