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Find the volume in the first octant bounded by the surfaces,
See figure attached for my attempt.
Not sure where I went wrong on this one, I think I got a good sketch of the actual volume, it looks like very thin slice of cheese of some sort.
Is the mistake in the way I set up my integral? Or in evaluating the integral itself? Or both?
Thanks again!
You have a very good sketch of the volume. Your problem is in setting up the integral. Think about the region in the xy plane that is one bound on your solid. It's bounded by y = x/2, y = 2x, and by the intersection of the plane 4x + 4y + z = 16 with the xy plane. What does this region look like?
I can picture the region the the xy plane in my mind and I'm thinking I'm gonna switch my order of integration. The area I'd have to find the xy plane would require me to use 2 seperate double integrals, and that's not what I want.Quote:
Originally Posted by Ackbeet
If I project my solid back into the yz plane I'd have a simple triangle, formed by z=0, y=0 and 4y + z = 16. This would only require 1 double integral to compute the area. Then I take this area and multiply it by my first integration to get the volume.
Edit: Hmmm... Integrating in the x direction first gives me a simple region in the yz plane however the first integration is ugly.
I'll try it your way and Integrate in the z direction first and I'll look at my region and see how things go. I'll post my results.
Edit: Okay I've got my 2nd attempt posted. The answer is listed to be 128/9 so it seems I just missed 1 negative sign somewhere. Can anyone spot it? Actually, I found it. It was on the 2nd last line in the term farthest to the right. It should be (-2/9 * 8)
So you've solved it, then?