anonymous
  • anonymous
**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 x-axis.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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ash2326
  • ash2326
|dw:1328198267200:dw|we have \[y=\sqrt{x+1}\] at x= 1 y=\(\sqrt 2\) at x = 5 y=\(\sqrt 6\) let's draw the graph first
ash2326
  • ash2326
now 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
  • ash2326
|dw:1328198515872:dw|

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ash2326
  • ash2326
now when this curve is rotated about x axis , this rectangle will be like a dish with thickness dx and radius \((y-0)\)(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 \[V=\pi /2 [36-4]\] so Volume = \[16\pi\]
anonymous
  • anonymous
This is for surface area , not volume.. :/
ash2326
  • ash2326
this is volume, I'm sure, do you have the answer
anonymous
  • anonymous
Sorry, I meant my question is asking for surface area not voulme.
amistre64
  • amistre64
im curious; if you took volume with the radius difference of "1" the value would be surface area maybe?
amistre64
  • amistre64
at any rate: integrate sqrt(dy^2+dx^2) is length of a curve
amistre64
  • amistre64
y = sqrt( x + 1) from x = 1 to x = 5 around the x-axis. \[\int_{1}^{5}2pi\ f(x)\ dx\] would seem right to me but i think I might be off somewhere
amistre64
  • amistre64
\[\int_{1}^{5}2pi\ f(x)\sqrt{1+[f'(x)]^2}\ dx\] seems to be the appropriate model tho
anonymous
  • anonymous
yes this is the right formula amistre64
amistre64
  • amistre64
then 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
  • amistre64
some simplification needed perhaps :)
anonymous
  • anonymous
yep.. 4x =4 in the denom, and then maybe find a common denom under the sq rt?
amistre64
  • amistre64
\[2pi\int_{1}^{5}\sqrt{(x+1)(1+\frac{1}{4(x+1)}})dx\] \[2pi\int_{1}^{5}\sqrt{(x+1)(\frac{4x+4+1}{4(x+1)}})dx\] \[2pi\int_{1}^{5}\sqrt{\cancel{(x+1)}(\frac{4x+5}{4\cancel{(x+1)}}})\] \[2pi\int_{1}^{5}\sqrt{\frac{4x+5}{4}}dx\] \[2pi\int_{1}^{5}\frac{\sqrt{4x+5}}{2}dx\] \[pi\int_{1}^{5}\sqrt{4x+5}dx\]
amistre64
  • amistre64
this looks more doable id say :)
amistre64
  • amistre64
this site is super slow with internet explorer.
amistre64
  • amistre64
with 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
  • amistre64
almost right; my 4s are mixed top to bottom :)
amistre64
  • amistre64
\[4pi(\frac{2}{4*3}(4x+5)^{3/2})\]
anonymous
  • anonymous
so /I got 86.1131pi, let me see if that is right.. Sorry my computer is sooo slow in the library.
ash2326
  • ash2326
guys check this video, I used this method http://www.khanacademy.org/video/solid-of-revolution--part-1?topic=calculus-1

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