# Area between curves

• Oct 15th 2009, 01:57 AM
iExcavate
Area between curves
Hey all!

I have this problem, and am unable to comprehend what to do, I haven't stumbled upon a question like this before. If anyone can offer assistance, it would be much appreciated.

Consider the two curves $\displaystyle y = 4/x$ and $\displaystyle y = -x+5$
(a) Calculate the area enclosed by the two curves.
(b) Find the volume of the solid generated by rotating the region about the y-axis.

Thanks (Happy)
• Oct 15th 2009, 02:42 AM
iExcavate
For: Part (a) Calculate the area enclosed by the two curves:

I got the answer: $\displaystyle (13/2)-4ln|4|$, is this correct?

Thanks again.
• Oct 15th 2009, 04:38 AM
HallsofIvy
No, it is not. Since you did not show your work, I cannot say any more than that.

As for the volume formed by rotating the region, there are many ways of doing that. What methods do you know?
• Oct 15th 2009, 05:09 PM
iExcavate
I used the formula: $\displaystyle A = \int_1^4 Yt - Yb dx$

$\displaystyle A = \int_1^4 (-x+5) - (4/x) dx$

Have I perhaps placed the functions incorrectly, using this formula?

Following through:
$\displaystyle [(-(x^2)/2 + 5x - 4ln|x|]_1^4$

$\displaystyle [(-((4)^2)/2 + 5(4) - 4ln|(4)|] - [(-((1)^2)/2 + 5(1) - 4ln|1|]$

$\displaystyle [-8 + 20 - 4ln|(4)|] - [1/2 + 5 - 0]$

$\displaystyle 12 - 4ln|(4)| - 11/2$

Finally giving the result:
$\displaystyle 13/2 - 4ln|4|$

Where did I go wrong? (Crying)
• Oct 16th 2009, 02:46 AM
nthethem
area between curves
looks like you used a valid method. Below is part of your working ....

http://www.mathhelpforum.com/math-he...01363b03-1.gif

http://www.mathhelpforum.com/math-he...31689cab-1.gif

note the 1/2 in the last bracket must be negative. That way, the answer becomes 15/2 - ln16.

For the second part you may

use pie*int[f(x)-g(x)]^2]dx
• Oct 16th 2009, 02:54 AM
nthethem
area between curves
• Oct 16th 2009, 04:25 PM
iExcavate
Where did you get 15/2 from? Shouldn't it be 35/2? ($\displaystyle 12 + 11/2 = 35/2$)

-> Giving the result: $\displaystyle (35/2)-ln|256|$?
• Oct 16th 2009, 05:12 PM
skeeter
Quote:

Originally Posted by iExcavate
Where did you get 15/2 from? Shouldn't it be 35/2? ($\displaystyle 12 + 11/2 = 35/2$)
-> Giving the result: $\displaystyle (35/2)-ln|256|$?

...
• Oct 16th 2009, 05:55 PM
iExcavate
11/2 = 5.5

12 - 11/2 = 13/2 (what I initially had)
12 - 5.5 = 6.5 -> 13/2

Where did 15/2 come from?
• Oct 16th 2009, 06:22 PM
skeeter
picking it up from scratch ...

$\displaystyle \left[-\frac{x^2}{2} + 5x - 4\ln{x}\right]_1^4$

$\displaystyle \left[-8 + 20 - 4\ln{4}\right] - \left[-\frac{1}{2} + 5 - 4\ln{1}\right]$

$\displaystyle 12 - 4\ln{4} - \frac{9}{2} = \frac{15}{2} - 8\ln{2}$