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Thread: Line integral

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

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

    My method:

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

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

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

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

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

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

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

    For the first integral:

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

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

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

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

    So the two integrals are now:

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

    For the second integral:

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

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

    Which becomes:

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

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

    so substituting the limits gives:

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

    So my overall answer to the question is:

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

    Firstly, is this right? I have an $\displaystyle arcosh 0$ 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 $\displaystyle \int_C(x+y^2)$, where $\displaystyle C$ is the parabola $\displaystyle y=x^2$ in the plane $\displaystyle z=0$ connecting the points $\displaystyle (0,0,0)$ and $\displaystyle (2,4,0)$.

    My method:

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

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

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



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



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

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

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

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

    For the first integral:

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

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

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

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


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

    Tonio



    So the two integrals are now:

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

    For the second integral:

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

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

    Which becomes:

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

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

    so substituting the limits gives:

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

    So my overall answer to the question is:

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

    Firstly, is this right? I have an $\displaystyle arcosh 0$ 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 $\displaystyle \int_C(x+y^2)$, where $\displaystyle C$ is the parabola $\displaystyle y=x^2$ in the plane $\displaystyle z=0$ connecting the points $\displaystyle (0,0,0)$ and $\displaystyle (2,4,0)$.

    My method:

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

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

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

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

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

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

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

    For the first integral:

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

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

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

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

    So the two integrals are now:

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

    For the second integral:

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

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

    Which becomes:

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

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

    so substituting the limits gives:

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

    So my overall answer to the question is:

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

    Firstly, is this right? I have an $\displaystyle arcosh 0$ 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 $\displaystyle y=x^2$ from $\displaystyle \left(0,0,0\right)$ to $\displaystyle \left(2,4,0\right)$ , so why don't you choose $\displaystyle x$ as parameter, with $\displaystyle 0\leq x \leq 2$??

    This way you have to evaluate the integral

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

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

    My method:

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

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

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

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

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

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

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

    For the first integral:

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

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

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

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

    So the two integrals are now:

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

    For the second integral:

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

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

    Which becomes:

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

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

    so substituting the limits gives:

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

    So my overall answer to the question is:

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

    Firstly, is this right? I have an $\displaystyle arcosh 0$ 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 $\displaystyle (\sqrt{2}x)^2= \cosh 2u$ which isn't that bad, but it is slightly more difficult than $\displaystyle (4x)^2=\cosh 2u$.
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