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  1. #1
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    Rotated about the...

    Okay so I'm having trouble formulating this integral. The region is bounded by:

    {x}^{2}-{y}^{2}=7 and x=4
    rotated about: y=5

    I tried using shells.. It didn't really work for me

    Thanks,
    CC
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  2. #2
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    Quote Originally Posted by CalculusCrazed View Post
    Okay so I'm having trouble formulating this integral. The region is bounded by:

    {x}^{2}-{y}^{2}=7 and x=4
    rotated about: y=5

    I tried using shells.. It didn't really work for me

    Thanks,
    CC
    What about

    V=2\pi \int_{\sqrt{7}}^4 x \cdot {\color{red}2}\sqrt{x^2-7}\, dx=\ldots =2\cdot 18\pi=36\pi?
    (use substitution u := x^2-7 to solve this integral)
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  3. #3
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    Quote Originally Posted by CalculusCrazed View Post
    Okay so I'm having trouble formulating this integral. The region is bounded by:

    {x}^{2}-{y}^{2}=7 and x=4
    rotated about: y=5

    I tried using shells.. It didn't really work for me

    Thanks,
    CC
    washers w/r to x ...

    R = 5 + \sqrt{x^2-7}<br />

    r = 5 - \sqrt{x^2-7}

    R^2 - r^2 = 25 - (x^2 - 7) = 32 - x^2

    V = \pi \int_{\sqrt{7}}^4 32 - x^2 \, dx
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  4. #4
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    Okay so you did R meaning the radius of the big circle and r for the little. How did you find the bounds?
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  5. #5
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    Quote Originally Posted by CalculusCrazed View Post
    Okay so you did R meaning the radius of the big circle and r for the little. How did you find the bounds?
    I sketched a graph ...

    upper branch is y = \sqrt{x^2-7}

    lower branch is y = -\sqrt{x^2-7}

    R = 5 - lower branch

    r = 5 - upper branch
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  6. #6
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    Okay, so why did you do R^2 - r^2?
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  7. #7
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    Quote Originally Posted by CalculusCrazed View Post
    Okay, so why did you do R^2 - r^2?
    volume by the method of washers ... a washer is a disk with a hole in it.

    V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx


    The Washer Method for Solids of Revolution
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  8. #8
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    The little circle minus the big circle. Okay I guess I just don't get how you got root 7 to 4
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  9. #9
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    Quote Originally Posted by CalculusCrazed View Post
    The little circle minus the big circle. Okay I guess I just don't get how you got root 7 to 4
    x = \sqrt{7} is the vertex of the hyperbola at the left end of the defined region to be rotated.

    x = 4 is the right boundary of the region to be rotated.

    I recommend you meet with your instructor ... this is all very basic material which should have been covered in class.
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  10. #10
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    ...plug in zero for y in other words.

    Thank you for your concern.
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