NathanJHW
  • NathanJHW
The region R is bounded by the x-axis, y-axis, x = 3 and y = 1/sqrt(x+1).
Mathematics
katieb
  • katieb
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NathanJHW
  • NathanJHW
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.
perl
  • perl
first lets graph the region https://www.desmos.com/calculator/2bz6ctgloa
NathanJHW
  • NathanJHW
Yeah I was able to do that much.

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perl
  • perl
so the area of the region R , we take the integral $$ \Large \int_{0}^{3} \frac{1}{\sqrt{x+1}}~dx$$
NathanJHW
  • NathanJHW
I got 2
perl
  • perl
thats correct so far
NathanJHW
  • NathanJHW
So 2 is the answer to A then right?
perl
  • perl
right
NathanJHW
  • NathanJHW
Alright, then how do we do part B?
perl
  • perl
part B we revolve about x axis $$ \Large \int_{0}^{3} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx$$
NathanJHW
  • NathanJHW
pi ln (4)
perl
  • perl
thats correct
NathanJHW
  • NathanJHW
Now how do we do part c?
perl
  • perl
we need to find the x point that divides the volume in half
perl
  • perl
|dw:1433367290427:dw|
perl
  • perl
you can't just cut it in half, because the region is not symmetric
NathanJHW
  • NathanJHW
So then where do we cut it?
perl
  • perl
solve for c $$\LARGE { \int_{0}^{\color{red}c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx = \frac{1}{2}\cdot \pi \ln 4 } $$
NathanJHW
  • NathanJHW
pi ln (c+1)
NathanJHW
  • NathanJHW
Wait sorry I'm solving for c not integrating.
NathanJHW
  • NathanJHW
I got ln|x+1| - ln|1| = ln|4| - ln|x+1|
NathanJHW
  • NathanJHW
x=1
perl
  • perl
right, so we need to solve ln| c + 1 | - ln| 0 + 1| = 1/2 * Pi * ln(4)
NathanJHW
  • NathanJHW
is it not x/c =1?
NathanJHW
  • NathanJHW
x/c means x or c as I put x when I solved.
perl
  • perl
$$ \Large { \int_{0}^{c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx \\~\\=\int_{0}^{c} \pi \left( \frac{1}{x+1} \right)~dx \\~\\ =\pi \left[ \ln |x + 1| \right]_{0}^{c} \\~\\ =\pi ( \ln|c+1| - \ln|0+1|) \\~\\ =\pi ( \ln|c+1| - \ln|1|) \\~\\ =\pi \ln|c+1| } $$
perl
  • perl
ok so far?
NathanJHW
  • NathanJHW
Yeah
perl
  • 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
NathanJHW
  • NathanJHW
Alright so I guess I was right, just needed to see the work.
NathanJHW
  • NathanJHW
Thanks for the help.
perl
  • perl
your welcome :)

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