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Math Help - Line integral

  1. #1
    Super Member Showcase_22's Avatar
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    Line integral

    Evaluate the line integral \int_C(x+y^2), where C is the parabola y=x^2 in the plane z=0 connecting the points (0,0,0) and (2,4,0).

    My method:

    Okay, so it's probably good to start with a parameterisation:

    \gamma (t)=(2t, 4t^2,0) where t \in [0,1].

    I need to find \int_0^1 f(\gamma (t)) || \gamma (t) ||~dt where f(x,y)=x+y^2

    \int_0^1 (2t+(4t^2)^2)(2^2+(8t^2))^{\frac{1}{2}}~dt

    \int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt

    2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt

    4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the first integral:

    Let u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t

    4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du

    \frac{1}{8} \left[ \frac{3}{2} u^{\frac{2}{3}} \right]_1^{17}

    =\frac{3}{16}(17^{\frac{2}{3}}-1)

    So the two integrals are now:

    \frac{3}{16}(17^{\frac{2}{3}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the second integral:

    Let (4t)^2=\cosh 2u

    Then, for the second integral, we have 32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du

    Which becomes:

    \frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du

    by FTC, the integral is \left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}

    so substituting the limits gives:

    \frac{1}{18} (16)^{\frac{9}{2}}

    So my overall answer to the question is:

    \frac{3}{16}(17^{\frac{2}{3}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}

    Firstly, is this right? I have an arcosh 0 in there that is undefined, but it removes itself quite nicely at the end.

    Secondly, is there a faster way of doing this???
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    Evaluate the line integral \int_C(x+y^2), where C is the parabola y=x^2 in the plane z=0 connecting the points (0,0,0) and (2,4,0).

    My method:

    Okay, so it's probably good to start with a parameterisation:

    \gamma (t)=(2t, 4t^2,0) where t \in [0,1].

    I need to find \int_0^1 f(\gamma (t)) || \gamma (t) ||~dt where f(x,y)=x+y^2



    == Probably a typo: if must be || \gamma^{'} (t)|| in the integral



    \int_0^1 (2t+(4t^2)^2)(2^2+(8t^2))^{\frac{1}{2}}~dt

    \int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt

    2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt

    4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the first integral:

    Let u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t

    4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du

    \frac{1}{8} \left[ \frac{3}{2} u^{\frac{2}{3}} \right]_1^{17}

    =\frac{3}{16}(17^{\frac{2}{3}}-1)


    === This is wrong: \int{\sqrt{u} \, du} = \frac{2}{3}u^{\frac{3}{2}} and not what you wrote

    Tonio



    So the two integrals are now:

    \frac{3}{16}(17^{\frac{2}{3}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the second integral:

    Let (4t)^2=\cosh 2u

    Then, for the second integral, we have 32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du

    Which becomes:

    \frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du

    by FTC, the integral is \left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}

    so substituting the limits gives:

    \frac{1}{18} (16)^{\frac{9}{2}}

    So my overall answer to the question is:

    \frac{3}{16}(17^{\frac{2}{3}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}

    Firstly, is this right? I have an arcosh 0 in there that is undefined, but it removes itself quite nicely at the end.

    Secondly, is there a faster way of doing this???

    I inserted a couple of times in your text above

    Tonio
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  3. #3
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    Evaluate the line integral \int_C(x+y^2), where C is the parabola y=x^2 in the plane z=0 connecting the points (0,0,0) and (2,4,0).

    My method:

    Okay, so it's probably good to start with a parameterisation:

    \gamma (t)=(2t, 4t^2,0) where t \in [0,1].

    I need to find \int_0^1 f(\gamma (t)) || \gamma \ ' (t) ||~dt where f(x,y)=x+y^2

    \int_0^1 (2t+(4t^2)^2)(2^2+(8t^2))^{\frac{1}{2}}~dt

    \int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt

    2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt

    4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the first integral:

    Let u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t

    4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du

    \frac{1}{8} \left[ \frac{2}{3} u^{\frac{2}{3}} \right]_1^{17}

    =\frac{1}{12}(17^{\frac{2}{3}}-1)

    So the two integrals are now:

    \frac{1}{12}(17^{\frac{2}{3}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the second integral:

    Let (4t)^2=\cosh 2u

    Then, for the second integral, we have 32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du

    Which becomes:

    \frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du

    by FTC, the integral is \left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}

    so substituting the limits gives:

    \frac{1}{18} (16)^{\frac{9}{2}}

    So my overall answer to the question is:

    \frac{1}{12}(17^{\frac{2}{3}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}

    Firstly, is this right? I have an arcosh 0 in there that is undefined, but it removes itself quite nicely at the end.

    Secondly, is there a faster way of doing this???
    eep! Both of those were typos! I remember seeing the second mistake you pointed out and thinking "gotta change that".

    *facepalm*

    But, overall, is this right now?
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    eep! Both of those were typos! I remember seeing the second mistake you pointed out and thinking "gotta change that".

    *facepalm*

    But, overall, is this right now?

    No, it isn't imo: you haven't yet changed the power of 17 in your answer!

    Why don't you do it the simple way: we're gonna go on y=x^2 from \left(0,0,0\right) to \left(2,4,0\right) , so why don't you choose x as parameter, with 0\leq x \leq 2??

    This way you have to evaluate the integral

    \int_0^2\left(x+x^4\right)\sqrt{\left(1+2x^2\right  )}~dx

    Tonio
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  5. #5
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    Evaluate the line integral \int_C(x+y^2), where C is the parabola y=x^2 in the plane z=0 connecting the points (0,0,0) and (2,4,0).

    My method:

    Okay, so it's probably good to start with a parameterisation:

    \gamma (t)=(2t, 4t^2,0) where t \in [0,1].

    I need to find \int_0^1 f(\gamma (t)) || \gamma \ ' (t) ||~dt where f(x,y)=x+y^2

    \int_0^1 (2t+(4t^2)^2)(2^2+(8t)^2)^{\frac{1}{2}}~dt

    \int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt

    2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt

    4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the first integral:

    Let u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t

    4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du

    \frac{1}{8} \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_1^{17}

    =\frac{1}{12}(17^{\frac{3}{2}}-1)

    So the two integrals are now:

    \frac{1}{12}(17^{\frac{3}{2}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt

    For the second integral:

    Let (4t)^2=\cosh 2u

    Then, for the second integral, we have 32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du

    Which becomes:

    \frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du

    by FTC, the integral is \left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}

    so substituting the limits gives:

    \frac{1}{18} (16)^{\frac{9}{2}}

    So my overall answer to the question is:

    \frac{1}{12}(17^{\frac{3}{2}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}

    Firstly, is this right? I have an arcosh 0 in there that is undefined, but it removes itself quite nicely at the end.

    Secondly, is there a faster way of doing this???
    Right, it's been amended.

    I don't think it's much easier parameterising between 0 and 2 since my substitution is now (\sqrt{2}x)^2= \cosh 2u which isn't that bad, but it is slightly more difficult than (4x)^2=\cosh 2u.
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