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Math Help - Calc 2 Problems

  1. #1
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    Calc 2 Problems

    1.) Set up the integral to find length of curve given y = \sqrt{x}, x =0, x=4

    2.) Set up the integral to find volume if \int_{0}^{4} \sqrt{x}\, dx is rotated around the x-axis.

    2.) Set up the integral to find volume if \int_{0}^{4} \sqrt{x}\, dx is rotated around the y-axis.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Ideasman View Post
    1.) Set up the integral to find length of curve given y = \sqrt{x}, x =0, x=4

    2.) Set up the integral to find volume if \int_{0}^{4} \sqrt{x}\, dx is rotated around the x-axis.

    2.) Set up the integral to find volume if \int_{0}^{4} \sqrt{x}\, dx is rotated around the y-axis.
    1) you are trying to find arclength so you appy the formula \int_0^4{\sqrt{1+\bigg(\frac{1}{2\sqrt{x}}\bigg)^2  }}dx

    for 2. I assume this is what you mean...from 0 to 4 with \sqrt{x}...so for x-axis we apply the formula to get {\pi}\int_0^4{x}dx

    and for 3. we apply the shell method formula to attain {2\pi}\int_0^4{x\cdot{\sqrt{x}}}dx
    Last edited by Mathstud28; April 7th 2008 at 02:02 PM.
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    1) you are trying to find arclength so you appy the formula \int_0^4{\sqrt{1+\bigg(\frac{1}{2\sqrt{x}}\bigg)^2  }}dx

    for 2. I assume this is what you mean...from 0 to 4 with \sqrt{x}...so for x-axis we apply the formula to get {\pi}\int_0^4{x}dx

    and for 3. we apply the shell method formula to attain {2\pi}\int_0^4{x\cdot{\sqrt{x}}}dx
    Sorry mistyped the question. Originally, the question was:

    1.) Compute the area bounded by y=\sqrt{x}, y= 0 and  x=4.

    Then, for 2 and 3 it was

    2.) Set up the integral to find the volume if this AREA is rotated about the x-axis

    3.) Set up the integral to find the volume if this AREA is rotated about the y-axis

    This answer still apply? I think 1 will be different, as what you gave me for #1 will be the length of the curve (arclength)

    Thanks a lot
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Ideasman View Post
    Sorry mistyped the question. Originally, the question was:

    1.) Compute the area bounded by y=\sqrt{x}, y= 0 and  x=4.

    Then, for 2 and 3 it was

    2.) Set up the integral to find the volume if this AREA is rotated about the x-axis

    3.) Set up the integral to find the volume if this AREA is rotated about the y-axis

    This answer still apply? I think 1 will be different, as what you gave me for #1 will be the length of the curve (arclength)

    Thanks a lot

    THe only one that changes is the first one it just becomes \int_0^4{\sqrt{x}}dx
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    for 2. I assume this is what you mean...from 0 to 4 with \sqrt{x}...so for x-axis we apply the formula to get {\pi}\int_0^4{x}dx
    Thanks a lot. Final question, for this, did you mean \sqrt{x} or simply  x?

    Edit: Oh I see why its just x.
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