View Full Version : Question about May 1992, Problem 12

robertr24

August 5th 2005, 04:26 PM

Hey all. I'm having a bit of a problem. In Dr. Ostazewski's manual, it's number 13 in practice exam 5. I'm confused about the area of integration. Since y >= x, I thought the area should have been x: (0 to w), y: (x to 1), and for the second part x: (w to sqrt w), y: (x to w/x)? In simpler terms, I thought the shaded region should be to the upper left of the line y = x instead of to the lower right. I don't get it. I would be more than grateful if someone could help me out. Thanks very much all.

krzysio

August 6th 2005, 12:44 AM

Dear Robertr24:

I do not know which edition of my manual you have: first or second edition, but

I think this is probably a typo that I have corrected in one of these posted errata:

First edition:

http://www.math.ilstu.edu/krzysio/P-Manual-Errata.pdf

Second edition:

http://www.math.ilstu.edu/krzysio/P-Manual-2ndEd-Errata.pdf

I hope this helps. If it does not, please reply, or send me an e-mail.

Yours most sincerely,

Krzys' Ostaszewski

Hey all. I'm having a bit of a problem. In Dr. Ostazewski's manual, it's number 13 in practice exam 5. I'm confused about the area of integration. Since y >= x, I thought the area should have been x: (0 to w), y: (x to 1), and for the second part x: (w to sqrt w), y: (x to w/x)? In simpler terms, I thought the shaded region should be to the upper left of the line y = x instead of to the lower right. I don't get it. I would be more than grateful if someone could help me out. Thanks very much all.

robertr24

August 6th 2005, 02:01 AM

Thanks. That clears things up a bit. However, I'm still confused about the actual integration. In the manual you have x:={0 to sqrt w}, y:={0 to x} for the first part, x:={sqrt w to 1}, y:={0 to x/w} for the second part. Doesn't this correspond to the area below the line y = x?

krzysio

August 6th 2005, 03:35 AM

The intersection of the line y = x and the hyperbola xy = w is at the point

( w^0.5, w^.05). First part is the integral over the shaded region between

the origin and the line y = w^.05, the second one is above that line

and still under the hyperbola. I still had typos in this solution, so I updated

the errata, and you can download corrected versions here:

http://www.math.ilstu.edu/krzysio/P-Manual-2ndEd-Errata.pdf

http://www.math.ilstu.edu/krzysio/P-Manual-Errata.pdf

Yours,

Krzys'

In the first part, we have

Thanks. That clears things up a bit. However, I'm still confused about the actual integration. In the manual you have x:={0 to sqrt w}, y:={0 to x} for the first part, x:={sqrt w to 1}, y:={0 to x/w} for the second part. Doesn't this correspond to the area below the line y = x?

robertr24

August 6th 2005, 12:17 PM

Thanks very much. I knew something was wrong there. It makes much more sense with the dx's and dy's interchanged. Sorry to drive you crazy. I will definitely look through ALL the errata before I post questions from now on. :)

krzysio

August 6th 2005, 12:39 PM

Dear Robert24:

Unfortunately, errors and typos still show up in my manuals, despite my best efforts.

So, I view your inquiry as a very nice gesture, helping me correct those errors. I appreciate it very much.

Yours very truly,

Krzys' Ostaszewski

Thanks very much. I knew something was wrong there. It makes much more sense with the dx's and dy's interchanged. Sorry to drive you crazy. I will definitely look through ALL the errata before I post questions from now on. :)

Powered by vBulletin® Version 4.2.2 Copyright © 2019 vBulletin Solutions, Inc. All rights reserved.