1. ## 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 $y = 4/x$ and $y = -x+5$
(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

2. For: Part (a) Calculate the area enclosed by the two curves:

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

Thanks again.

3. 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?

4. I used the formula: $A = \int_1^4 Yt - Yb dx$

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

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

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

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

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

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

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

Where did I go wrong?

5. ## area between curves

looks like you used a valid method. Below is part of your working ....

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

6. ## area between curves

7. Where did you get 15/2 from? Shouldn't it be 35/2? ( $12 + 11/2 = 35/2$)

-> Giving the result: $(35/2)-ln|256|$?

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

9. 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?

10. picking it up from scratch ...

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

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

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