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anonymous
 4 years ago
**Calc 2 help needed** Find the area of the surface generated by revolving the curve y = sqrt( x + 1) from x = 1 to x = 5 around the xaxis.
anonymous
 4 years ago
**Calc 2 help needed** Find the area of the surface generated by revolving the curve y = sqrt( x + 1) from x = 1 to x = 5 around the xaxis.

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ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328198267200:dwwe have \[y=\sqrt{x+1}\] at x= 1 y=\(\sqrt 2\) at x = 5 y=\(\sqrt 6\) let's draw the graph first

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0now for finding the volume let's take a recatngle with length length y and width dx if we were to find the area \[A=\int_{x=1}^{x=5} (\sqrt{x+1}0) dx\]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0now when this curve is rotated about x axis , this rectangle will be like a dish with thickness dx and radius \((y0)\)(length of the rectangle) so volume of this dish with radius y and thickness dx is \(\pi*y^2*dx\) so volume of the whole curve rotated around x axis \[V=\int_{x=1}^{x=5} \pi*(y)^2*dx\] y=(x+1)^1/2 so \[V=\int_{x=1}^{x=5} \pi*(x+1)*dx\] or \[ V= \pi(x+1)^2 /2\] insert the limits x=1 to x=5 \[V= \pi /2 [(5+1)^2(1+1)^2]\] or \[V=\pi /2 [364]\] so Volume = \[16\pi\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is for surface area , not volume.. :/

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0this is volume, I'm sure, do you have the answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, I meant my question is asking for surface area not voulme.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0im curious; if you took volume with the radius difference of "1" the value would be surface area maybe?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0at any rate: integrate sqrt(dy^2+dx^2) is length of a curve

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0y = sqrt( x + 1) from x = 1 to x = 5 around the xaxis. \[\int_{1}^{5}2pi\ f(x)\ dx\] would seem right to me but i think I might be off somewhere

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_{1}^{5}2pi\ f(x)\sqrt{1+[f'(x)]^2}\ dx\] seems to be the appropriate model tho

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes this is the right formula amistre64

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0then lets plug in parts :) \[ y = \sqrt{ x + 1}\] \[ y' = \frac{1}{2\sqrt{ x + 1}}\] \[2pi\int_{1}^{5}\sqrt{x+1}\sqrt{1+\frac{1}{4(x+1)}}dx\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0some simplification needed perhaps :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep.. 4x =4 in the denom, and then maybe find a common denom under the sq rt?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[2pi\int_{1}^{5}\sqrt{(x+1)(1+\frac{1}{4(x+1)}})dx\] \[2pi\int_{1}^{5}\sqrt{(x+1)(\frac{4x+4+1}{4(x+1)}})dx\] \[2pi\int_{1}^{5}\sqrt{\cancel{(x+1)}(\frac{4x+5}{4\cancel{(x+1)}}})\] \[2pi\int_{1}^{5}\sqrt{\frac{4x+5}{4}}dx\] \[2pi\int_{1}^{5}\frac{\sqrt{4x+5}}{2}dx\] \[pi\int_{1}^{5}\sqrt{4x+5}dx\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0this looks more doable id say :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0this site is super slow with internet explorer.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0with the 4x under the sqrt; it tells me we need to borrow a 4 from a useful "1"; say 4/4 \[\frac{pi}{4}\int_{1}^{5}4(4x+5)^{1/2}dx\] \[\frac{pi}{4}(\frac{4*2}{3}(4x+5)^{3/2})\] ::at 5 and 1

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0almost right; my 4s are mixed top to bottom :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[4pi(\frac{2}{4*3}(4x+5)^{3/2})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so /I got 86.1131pi, let me see if that is right.. Sorry my computer is sooo slow in the library.

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0guys check this video, I used this method http://www.khanacademy.org/video/solidofrevolutionpart1?topic=calculus1
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