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anonymous

  • 4 years ago

**Calc2 Help Needed** Find the surface area of the curve y= sqrt(x + 1) from x = 1 to x = 5 around the x-axis.

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  1. anonymous
    • 4 years ago
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    I am getting 196/3 PI as my final answer.

  2. anonymous
    • 4 years ago
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    \[2\pi \int\limits \sqrt(x+1) * \sqrt( 1 + (1/(2\sqrt(x+1)))^2\]

  3. anonymous
    • 4 years ago
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    Someone help... Ahhh.. haha

  4. TuringTest
    • 4 years ago
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    I don't like it from there so I am about to do this in terms of y I am hopeful that will be prettier

  5. TuringTest
    • 4 years ago
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    yeah, definitely doable in terms of y.

  6. TuringTest
    • 4 years ago
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    \[y=\sqrt{x+1}\to x=y^2-1\]\[\frac{dx}{dy}=2y\]\[ds=\sqrt{1+(2y)^2}dy\]\[S=2\pi\int_{\sqrt2}^{\sqrt6} yds=2\pi\int_{\sqrt2}^{\sqrt6}y\sqrt{1+4y^2}dy\]\[u=1+4y^2\to du=8ydy\]so the integral is now\[\frac{\pi}4\int\sqrt udu\]

  7. anonymous
    • 4 years ago
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    which is 2/3 u^3/2..

  8. TuringTest
    • 4 years ago
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    ...times pi/4 change back to y evaluate at limits

  9. anonymous
    • 4 years ago
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    Do i not need to change the limits since u-sub?

  10. TuringTest
    • 4 years ago
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    I would just change this back to terms of y before evaluating

  11. TuringTest
    • 4 years ago
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    2/3(1+4y^2)^(3/2) evaluated at sqrt2 to sqrt6

  12. anonymous
    • 4 years ago
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    so 3/2 x = y^3/2

  13. anonymous
    • 4 years ago
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    raise both sides to 2/3?

  14. TuringTest
    • 4 years ago
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    I'm not sure what you mean...

  15. anonymous
    • 4 years ago
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    would it be y = (3/2 *x)^2/3?

  16. TuringTest
    • 4 years ago
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    \[y=\sqrt{x+1}\to x=y^2-1\]\[\frac{dx}{dy}=2y\]\[ds=\sqrt{1+(2y)^2}dy\]\[S=2\pi\int_{\sqrt2}^{\sqrt6} yds=2\pi\int_{\sqrt2}^{\sqrt6}y\sqrt{1+4y^2}dy\]\[u=1+4y^2\to du=8ydy\]so the integral is now\[\frac{\pi}4\int_{a}^{b}\sqrt udu=\frac{\pi}6u^{3/2}=\frac{\pi}6(1+4y^2)|_{\sqrt2}^{\sqrt6}\]I don't see why you still have both y and x in what you are doing...

  17. TuringTest
    • 4 years ago
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    where do you get 3/2x ?

  18. TuringTest
    • 4 years ago
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    Anybody reading this if you could take a look at my solution here and find my error it would be greatly appreciated! http://openstudy.com/study#/updates/4f2b199ce4b039c5a5c85318

  19. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=integrate+pi%28sqrt%284x%2B5%29%29+from+1+to+5

  20. amistre64
    • 4 years ago
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    im assuming this is the same one as before ..

  21. TuringTest
    • 4 years ago
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    haha, I was trippin' earlier :D I could have done it just fine in x, but my brain wasn't working in the simplification

  22. amistre64
    • 4 years ago
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    \[\int 2pi\ f(x)\sqrt{1+[f'(x)]^2}dxright?\]

  23. anonymous
    • 4 years ago
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    integral of sqrt(4x + 5) is where I got to.

  24. anonymous
    • 4 years ago
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    Yep that is right.

  25. TuringTest
    • 4 years ago
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    hey amistre if you could have a look at the one in the link I posted that would be cool

  26. amistre64
    • 4 years ago
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    the link aint taking me anywhere

  27. TuringTest
    • 4 years ago
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    ...by the way my answer is right as well, just for the record ;) http://www.wolframalpha.com/input/?i=integral%202pi%20y*sqrt(1%2B4y%5E2)dy%20from%20sqrt2%20to%20sqrt%206&t=crmtb01 @amistre then look at the last question asked by Cameron

  28. anonymous
    • 4 years ago
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    Oh wow 49/3 pi is right.. This thing is just really picky about decimals!

  29. amistre64
    • 4 years ago
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    just one of those days ..... still cant get a bead on your link

  30. anonymous
    • 4 years ago
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    http://openstudy.com/users/cameronmx9#/updates/4f2aaff5e4b049df4e9e3d87

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