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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 x-axis.

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  1. ash2326
    • 4 years ago
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    |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

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

  3. ash2326
    • 4 years ago
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    |dw:1328198515872:dw|

  4. ash2326
    • 4 years ago
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    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 = \[16\pi\]

  5. anonymous
    • 4 years ago
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    This is for surface area , not volume.. :/

  6. ash2326
    • 4 years ago
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    this is volume, I'm sure, do you have the answer

  7. anonymous
    • 4 years ago
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    Sorry, I meant my question is asking for surface area not voulme.

  8. amistre64
    • 4 years ago
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    im curious; if you took volume with the radius difference of "1" the value would be surface area maybe?

  9. amistre64
    • 4 years ago
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    at any rate: integrate sqrt(dy^2+dx^2) is length of a curve

  10. amistre64
    • 4 years ago
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    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

  11. amistre64
    • 4 years ago
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    \[\int_{1}^{5}2pi\ f(x)\sqrt{1+[f'(x)]^2}\ dx\] seems to be the appropriate model tho

  12. anonymous
    • 4 years ago
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    yes this is the right formula amistre64

  13. amistre64
    • 4 years ago
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    then lets plug in parts :) \[ y = \sqrt{ x + 1}\] \[ y' = \frac{1}{2\sqrt{ x + 1}}\] \[2pi\int_{1}^{5}\sqrt{x+1}\sqrt{1+\frac{1}{4(x+1)}}dx\]

  14. amistre64
    • 4 years ago
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    some simplification needed perhaps :)

  15. anonymous
    • 4 years ago
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    yep.. 4x =4 in the denom, and then maybe find a common denom under the sq rt?

  16. amistre64
    • 4 years ago
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    \[2pi\int_{1}^{5}\sqrt{(x+1)(1+\frac{1}{4(x+1)}})dx\] \[2pi\int_{1}^{5}\sqrt{(x+1)(\frac{4x+4+1}{4(x+1)}})dx\] \[2pi\int_{1}^{5}\sqrt{\cancel{(x+1)}(\frac{4x+5}{4\cancel{(x+1)}}})\] \[2pi\int_{1}^{5}\sqrt{\frac{4x+5}{4}}dx\] \[2pi\int_{1}^{5}\frac{\sqrt{4x+5}}{2}dx\] \[pi\int_{1}^{5}\sqrt{4x+5}dx\]

  17. amistre64
    • 4 years ago
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    this looks more doable id say :)

  18. amistre64
    • 4 years ago
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    this site is super slow with internet explorer.

  19. amistre64
    • 4 years ago
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    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

  20. amistre64
    • 4 years ago
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    almost right; my 4s are mixed top to bottom :)

  21. amistre64
    • 4 years ago
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    \[4pi(\frac{2}{4*3}(4x+5)^{3/2})\]

  22. anonymous
    • 4 years ago
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    so /I got 86.1131pi, let me see if that is right.. Sorry my computer is sooo slow in the library.

  23. ash2326
    • 4 years ago
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    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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