# [SOLVED] Volume by integration not quite working out

• Apr 13th 2009, 04:15 PM
Grep
[SOLVED] Volume by integration not quite working out
Hi all, my first post to this fine forum. I've been studying (on my own) and I'm doing volumes by integration at the moment. I've just hit a problem that doesn't seem right. I hesitate to wonder if there's an error in the textbook, but I've found a number of them so far (they should seriously publish errata for these books!), so perhaps... Help greatly appreciated here.

I'm using Basic Technical Mathematics with Calculus 6th edition, from Addison Wesley. Page 738, exercise 19. I'm supposed to be revolving them around the y-axis and use the indicated method. The problem is:

$\displaystyle x^2 - 4y^2 = 4, x=3$ (shells)

So I need to use the shells method. Ok, it's obviously a hyperbola. It crosses the x-axis at x=2. So I find radius = x, height = y, thickness = dx.

Solving for y gives me: $\displaystyle y = \sqrt{\frac{1} {4}x^2 - 1}$

Putting together the integral:

$\displaystyle V = 2\pi \int_2^3 xy ~dx = 2\pi \int_2^3 (\frac {1} {4} x^2 - 1)^\frac {1} {2} ~x ~dx = \frac {8\pi}{3}(\frac {1}{4} x^2 - 1)^\frac {3}{2}$

Evaluating that, I get $\displaystyle 3.727\pi$. The textbook gives an answer of $\displaystyle \frac {10\pi}{3}\sqrt{5}$ which works out in decimal to $\displaystyle 7.454\pi$, which is pretty much double the answer I get.

Odds are I'm the one that made the mistake... lol... But I just can't seem to get the textbook answer. Hope I got all that nicely formatted so it's easy to read. First time using LaTex and all.

And thanks again for all the valuable information in this forum, and for all the help that is given!

Grep.
• Apr 13th 2009, 04:45 PM
skeeter
Quote:

Originally Posted by Grep
Hi all, my first post to this fine forum. I've been studying (on my own) and I'm doing volumes by integration at the moment. I've just hit a problem that doesn't seem right. I hesitate to wonder if there's an error in the textbook, but I've found a number of them so far (they should seriously publish errata for these books!), so perhaps... Help greatly appreciated here.

I'm using Basic Technical Mathematics with Calculus 6th edition, from Addison Wesley. Page 738, exercise 19. I'm supposed to be revolving them around the y-axis and use the indicated method. The problem is:

$\displaystyle x^2 - 4y^2 = 4, x=3$ (shells)

So I need to use the shells method. Ok, it's obviously a hyperbola. It crosses the x-axis at x=2. So I find radius = x, height = y, thickness = dx.

Solving for y gives me: $\displaystyle y = \sqrt{\frac{1} {4}x^2 - 1}$

Putting together the integral:

$\displaystyle V = 2\pi \int_2^3 xy ~dx = 2\pi \int_2^3 (\frac {1} {4} x^2 - 1)^\frac {1} {2} ~x ~dx = \frac {8\pi}{3}(\frac {1}{4} x^2 - 1)^\frac {3}{2}$

Evaluating that, I get $\displaystyle 3.727\pi$. The textbook gives an answer of $\displaystyle \frac {10\pi}{3}\sqrt{5}$ which works out in decimal to $\displaystyle 7.454\pi$, which is pretty much double the answer I get.

Odds are I'm the one that made the mistake... lol... But I just can't seem to get the textbook answer. Hope I got all that nicely formatted so it's easy to read. First time using LaTex and all.

And thanks again for all the valuable information in this forum, and for all the help that is given!

Grep.

the region in question goes above and below the x-axis ... you only rotated the upper half.
• Apr 13th 2009, 04:51 PM
Grep
LOL That would explain why my answer is half the one in the textbook, wouldn't it? /facepalm

Thanks a lot skeeter, I very much appreciate your helping me out. And I can make sure I don't repeat the mistake in the future. :)

Grep.
• Apr 13th 2009, 05:00 PM
Reckoner
Just to narrow down your mistake, in case you haven't realized it yet:

Quote:

Originally Posted by Grep
Solving for y gives me: $\displaystyle y = \sqrt{\frac{1} {4}x^2 - 1}$

Solving for $\displaystyle y$ actually gives

$\displaystyle y=\color{red}\pm\color{black}\sqrt{\frac{x^2}4-1}.$

In general, the solution to $\displaystyle x^2=a$ is $\displaystyle x=\pm\sqrt a$ (for $\displaystyle a\geq0$). Why? Because the square of a number is the same as the square of its negative. So if, e.g., $\displaystyle x^2=4,\;x$ could be 2, but it could also be -2 since $\displaystyle (-2)^2=4.$

So, the height of your shells is actually

$\displaystyle \text{upper curve}-\text{lower curve}$

$\displaystyle =\left(\sqrt{\frac{x^2}4-1}\right)-\left(-\sqrt{\frac{x^2}4-1}\right)$

$\displaystyle =\sqrt{\frac{x^2}4-1}+\sqrt{\frac{x^2}4-1}$

$\displaystyle =2\sqrt{\frac{x^2}4-1}.$

Edit: And congratulations on your decision to self-teach. I am sure that you will find it very rewarding.
• Apr 13th 2009, 05:27 PM
Grep
Thanks Reckoner. The square root thing I'd noticed, but I solved it again by using 2y as my height. Which works out to the same, of course, but I like the way you worked it out better (upper curve - lower curve), and it will also work if the top and bottom curves are different. So I'll approach it in the way you described next time, I think. It's like what I was doing with areas anyways.

And thanks for the self-teaching comment. :) I learn better from books than most teachers anyways. It's only when I get stuck on something that I really wish I had a teacher handy, and this forum does fill the gap quite well.

And I do love Calculus! A bit to go before I hit manifolds and differential geometry, though, alas... lol. General Relativity, you will be mine!

Thanks again for the insightful comments.

Grep.