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Math Help - Solving for the Area under a Curve rotated around the x-axis.

  1. #1
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    [SOLVED]Solving for the Area under a Curve rotated around the x-axis.

    I know how to use the shell method for this problem. But I'm not sure how to solve it using disk or washer. I know you solve for X which gives you X=(Y)^(.25).

    The parameters are. Y=x^4 x=0 y=81

    I know Solving for the Area under a Curve rotated around the x-axis.-untitled.png is the equation for washer method.

    This mean Solving for the Area under a Curve rotated around the x-axis.-untitled2.png. Now I'm lost? What's the Big R and Little r now.
    Big R= 81-0?
    Little r= (Y^(.25))-0?
    Last edited by Bracketology; November 7th 2010 at 05:36 PM. Reason: Solved.
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  2. #2
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    Quote Originally Posted by Bracketology View Post
    I know how to use the shell method for this problem. But I'm not sure how to solve it using disk or washer. I know you solve for X which gives you X=(Y)^(.25).

    The parameters are. Y=x^4 x=0 y=81

    I know Click image for larger version. 

Name:	Untitled.png 
Views:	21 
Size:	784 Bytes 
ID:	19625 is the equation for washer method.

    This mean Click image for larger version. 

Name:	Untitled2.png 
Views:	20 
Size:	858 Bytes 
ID:	19627. Now I'm lost? What's the Big R and Little r now.
    Big R= 81-0?
    Little r= (Y^(.25))-0?
    rotation of the region about the x or y axis?
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  3. #3
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    Around the x-axis. Would I use disk method?
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  4. #4
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    Quote Originally Posted by Bracketology View Post
    Around the x-axis. Would I use disk method?
    yes, disks ...

    \displaystyle V = \pi \int_0^3 81^2 - (x^4)^2 \, dx
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  5. #5
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    Oops....For some reason I thought the x-axis was the y-axis. Thanks for your help.
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