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Math Help - area of the surface obtained by rotating the curve

  1. #1
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    Post area of the surface obtained by rotating the curve

    4) Find the area of the surface obtained by rotating the curve
    from to about the -axis.

    Any idea about the answer? I try to do it and my answer always turn out wrong. I don't know why
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by nhatie View Post
    4) Find the area of the surface obtained by rotating the curve
    from to about the -axis.

    Any idea about the answer? I try to do it and my answer always turn out wrong. I don't know why
    S = \int_{a}^{b} 2\pi f(x) \sqrt{1+(f'(x))^2} dx around x-axis

    so we need to find dy/dx
    dy/dx = \frac{5}{2\sqrt{5x}}

    the surface area

    S = 2\pi \int_{0}^{4} \sqrt{5x} \sqrt{1+ \frac{25}{4(5x)}} dx
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  3. #3
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    Quote Originally Posted by Amer View Post
    S = \int_{a}^{b} 2\pi f(x) \sqrt{1+(f'(x))^2} dx around x-axis

    so we need to find dy/dx
    dy/dx = \frac{5}{2\sqrt{5x}}

    the surface area

    S = 2\pi \int_{0}^{4} \sqrt{5x} \sqrt{1+ \frac{25}{4(5x)}} dx
    I worked out to that part and got stuck there? How can you solve it?
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by nhatie View Post
    I worked out to that part and got stuck there? How can you solve it?
    I will not solve it

    S = 2\pi \int_0^4 \sqrt{5x + \frac{25}{4} } dx

    you should know how to integrate this
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  5. #5
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    Quote Originally Posted by Amer View Post
    I will not solve it

    S = 2\pi \int_0^4 \sqrt{5x + \frac{25}{4} } dx

    you should know how to integrate this
    i got : -2(5x +25/4)^-1/2 from 0 to 4 , is it rite??
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  6. #6
    MHF Contributor Amer's Avatar
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    drive -2(5x +25/4)^{-1/2} it should equal \sqrt{5x + \frac{25}{4} } if it is true
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  7. #7
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    Quote Originally Posted by nhatie View Post
    i got : -2(5x +25/4)^-1/2 from 0 to 4 , is it rite??
    Almost but not quite.

    Let u= 5x+ 25/4. Then du= 5dx so that dx= (1/5)du and the integral is (1/5)\int u^{1/2}du. You don't have the "1/5" and the integral of u^{1/2} is [tex](2/3)u^{3/2}[/itex]. It looks like you differentiated rather that integrating.
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    Almost but not quite.

    Let u= 5x+ 25/4. Then du= 5dx so that dx= (1/5)du and the integral is (1/5)\int u^{1/2}du. You don't have the "1/5" and the integral of u^{1/2} is [tex](2/3)u^{3/2}[/itex]. It looks like you differentiated rather that integrating.
    ok i have it now and i have this
    1/5 int 2/3u^(3/2) => 1/5(2/3(5x+25/4)^3/2) |from 0 to 4
    answer: 15.8! is this rite?
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  9. #9
    MHF Contributor Amer's Avatar
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    correct solution
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  10. #10
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    Quote Originally Posted by Amer View Post
    correct solution
    but when i put it in.. the answe is wrong? can you recaculate my work?
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  11. #11
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by nhatie View Post
    but when i put it in.. the answe is wrong? can you recaculate my work?
    multiply it with 2\pi
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  12. #12
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    Quote Originally Posted by Amer View Post
    multiply it with 2\pi
    i have 99.5 but it is still wrong!!
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