anonymous one year ago The region R is bounded by the x-axis, y-axis, x = 3 and y = 1/sqrt(x+1).

1. anonymous

A. Find the area of region R. B.Find the volume of the solid formed when the region R is revolved about the x-axis C. The solid formed in part B is divided into two solids of equal volume by a plane perpendicular to the x-axis. Find the x-value where this plane intersects the x-axis.

2. perl

first lets graph the region https://www.desmos.com/calculator/2bz6ctgloa

3. anonymous

Yeah I was able to do that much.

4. perl

so the area of the region R , we take the integral $$\Large \int_{0}^{3} \frac{1}{\sqrt{x+1}}~dx$$

5. anonymous

I got 2

6. perl

thats correct so far

7. anonymous

So 2 is the answer to A then right?

8. perl

right

9. anonymous

Alright, then how do we do part B?

10. perl

part B we revolve about x axis $$\Large \int_{0}^{3} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx$$

11. anonymous

pi ln (4)

12. perl

thats correct

13. anonymous

Now how do we do part c?

14. perl

we need to find the x point that divides the volume in half

15. perl

|dw:1433367290427:dw|

16. perl

you can't just cut it in half, because the region is not symmetric

17. anonymous

So then where do we cut it?

18. perl

solve for c $$\LARGE { \int_{0}^{\color{red}c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx = \frac{1}{2}\cdot \pi \ln 4 }$$

19. anonymous

pi ln (c+1)

20. anonymous

Wait sorry I'm solving for c not integrating.

21. anonymous

I got ln|x+1| - ln|1| = ln|4| - ln|x+1|

22. anonymous

x=1

23. perl

right, so we need to solve ln| c + 1 | - ln| 0 + 1| = 1/2 * Pi * ln(4)

24. anonymous

is it not x/c =1?

25. anonymous

x/c means x or c as I put x when I solved.

26. perl

$$\Large { \int_{0}^{c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx \\~\\=\int_{0}^{c} \pi \left( \frac{1}{x+1} \right)~dx \\~\\ =\pi \left[ \ln |x + 1| \right]_{0}^{c} \\~\\ =\pi ( \ln|c+1| - \ln|0+1|) \\~\\ =\pi ( \ln|c+1| - \ln|1|) \\~\\ =\pi \ln|c+1| }$$

27. perl

ok so far?

28. anonymous

Yeah

29. perl

so we need to solve pi ln ( c + 1) = 1/2 * pi * ln(4) ln(c+1) = 1/2 * ln (4) ln(c+1) = ln(4 ^(1/2) ) ln(c+1) = ln(2) so c = 1

30. anonymous

Alright so I guess I was right, just needed to see the work.

31. anonymous

Thanks for the help.

32. perl