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Math Help - find the volume of the solid by rotating..

  1. #1
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    find the volume of the solid by rotating..

    hi guys, having problems with this practice question.


    Find the volume of the solid obtained by rotating the region under the graph of y = \sqrt{x} between x = 2 and x = 7 around the x-axis.

    thank u for help


    EDIT: i rotated the graph of y = \sqrt{x} then i grabed the intervals 2 till 7. so is the integration \int_2^{7} (x^{1/2})^2
    Last edited by jvignacio; November 10th 2008 at 09:30 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jvignacio View Post
    hi guys, having problems with this practice question.


    Find the volume of the solid obtained by rotating the region under the graph of y = \sqrt{x} between x = 2 and x = 7 around the x-axis.

    thank u for help


    EDIT: i rotated the graph of y = \sqrt{x} then i grabed the intervals 2 till 7. so is the integration \int_2^{7} (x^{1/2})^2
    The volume of revolution can be envisaged as approximatly composed of disks of radius \sqrt{x} and thickness \delta x , and hence volume of each disk is \pi |x| \delta x. Then the volume of the solid of revolution is approximatly the sum of the volumes of the disks comprising the solid, or in the limit the integral:

    V=\int_2^7 \pi |x|\ dx

    and as x is positive over the range of integration we can drop the |.| to get:

    V=\int_2^7 \pi x\ dx

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    The volume of revolution can be envisaged as approximatly composed of disks of radius \sqrt{x} and thickness \delta x , and hence volume of each disk is \pi |x| \delta x. Then the volume of the solid of revolution is approximatly the sum of the volumes of the disks comprising the solid, or in the limit the integral:

    V=\int_2^7 \pi |x|\ dx

    and as x is positive over the range of integration we can drop the |.| to get:

    V=\int_2^7 \pi x\ dx

    CB
    so now i just integrate V=\int_2^7 \pi x\ dx ?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jvignacio View Post
    so now i just integrate V=\int_2^7 \pi x\ dx ?
    Yes.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Yes.

    CB
    V=\int_2^7 \pi x\ dx

    = 70.6858347 correct?
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  6. #6
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    Quote Originally Posted by jvignacio View Post
    V=\int_2^7 \pi x\ dx

    = 70.6858347 correct?
    You should leave it as \frac{45 \pi}{2}

    CB
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