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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A. Find the area of region R. B.Find the volume of the solid formed when the region R is revolved about the xaxis C. The solid formed in part B is divided into two solids of equal volume by a plane perpendicular to the xaxis. Find the xvalue where this plane intersects the xaxis.

perl
 one year ago
Best ResponseYou've already chosen the best response.1first lets graph the region https://www.desmos.com/calculator/2bz6ctgloa

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I was able to do that much.

perl
 one year ago
Best ResponseYou've already chosen the best response.1so the area of the region R , we take the integral $$ \Large \int_{0}^{3} \frac{1}{\sqrt{x+1}}~dx$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 2 is the answer to A then right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, then how do we do part B?

perl
 one year ago
Best ResponseYou've already chosen the best response.1part B we revolve about x axis $$ \Large \int_{0}^{3} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now how do we do part c?

perl
 one year ago
Best ResponseYou've already chosen the best response.1we need to find the x point that divides the volume in half

perl
 one year ago
Best ResponseYou've already chosen the best response.1you can't just cut it in half, because the region is not symmetric

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then where do we cut it?

perl
 one year ago
Best ResponseYou've already chosen the best response.1solve for c $$\LARGE { \int_{0}^{\color{red}c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx = \frac{1}{2}\cdot \pi \ln 4 } $$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait sorry I'm solving for c not integrating.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got lnx+1  ln1 = ln4  lnx+1

perl
 one year ago
Best ResponseYou've already chosen the best response.1right, so we need to solve ln c + 1   ln 0 + 1 = 1/2 * Pi * ln(4)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x/c means x or c as I put x when I solved.

perl
 one year ago
Best ResponseYou've already chosen the best response.1$$ \Large { \int_{0}^{c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx \\~\\=\int_{0}^{c} \pi \left( \frac{1}{x+1} \right)~dx \\~\\ =\pi \left[ \ln x + 1 \right]_{0}^{c} \\~\\ =\pi ( \lnc+1  \ln0+1) \\~\\ =\pi ( \lnc+1  \ln1) \\~\\ =\pi \lnc+1 } $$

perl
 one year ago
Best ResponseYou've already chosen the best response.1so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright so I guess I was right, just needed to see the work.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help.
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