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Math Help - How to evaluate this integral by hand?

  1. #1
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    Exclamation How to evaluate this integral by hand?

    Hello everyone,
    \[\int\limits_0^{\sqrt 2 h} {\frac{\tau }{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}d} \tau \]
    I want to evaluate the above integral by hand. It is from a question in Griffiths Introduction to Electrodynamics. Can you explain how to do step by step?

    P.S: I did it with Wolframalpha but failed to understand it since it gave the desired answer for restricted values of x and h values. What does that mean by the way?

    Thanks in advance
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  2. #2
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    Re: How to evaluate this integral by hand?

    Quote Originally Posted by JohnDoe View Post
    Hello everyone,
    \[\int\limits_0^{\sqrt 2 h} {\frac{\tau }{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}d} \tau \]
    I have not done the integral, but I suspect it involves:
    r^2  + h^2  - \sqrt 2 hr = \left( {r - \frac{h}{{\sqrt 2 }}} \right)^2  + \frac{{h^2 }}{2}~.
    Last edited by Plato; June 19th 2011 at 12:30 PM.
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  3. #3
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    Re: How to evaluate this integral by hand?

    +h/2 only
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  4. #4
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    Re: How to evaluate this integral by hand?

    Yes thanks for your help but I am stuck after these step, I cannot find the integral I might be doing something wrong here is my work:
    \frac{{\sigma 2\pi }}{{4\pi {\varepsilon _0}\sqrt 2 }}\int\limits_0^{\sqrt 2 h} {\frac{{\tau d\tau }}{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}} \\<br />
{\tau ^2} + {h^2} - \sqrt 2 h\tau  = {(\tau  - \frac{h}{{\sqrt 2 }})^2} + \frac{{{h^2}}}{2}\\<br />
u = (\tau  - \frac{h}{{\sqrt 2 }})\\<br />
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_a^b {\frac{{u + \frac{h}{{\sqrt 2 }}du}}{{\sqrt {{u^2} + \frac{{{h^2}}}{2}} }}} \\
    \frac{h}{{\sqrt 2 }}\tan \theta  = u\\<br />
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_c^d {\frac{{\frac{h}{{\sqrt 2 }}(1 + \tan \theta )\frac{h}{{\sqrt 2 }}{{\sec }^2}\theta d\theta }}{{\frac{h}{{\sqrt 2 }}\sec \theta }}} \\<br />
\frac{\sigma }{{\pi {\varepsilon _0}4}}\int\limits_c^d {(\sec \theta  + \sec \theta \tan \theta )d\theta }  = \frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{u^2} + {h^2}} }}{h} + \frac{{\sqrt 2 u}}{h}) + \frac{{\sqrt {2{u^2} + {h^2}} }}{h}]\\<br />
    \frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{{(\tau  - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} + \frac{{\sqrt 2 (\tau  - \frac{h}{{\sqrt 2 }})}}{h}) + \frac{{\sqrt {2{{(\tau  - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h}]_0^{\sqrt 2 h}\\<br />
\frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau }  + \sqrt 2 \tau  - h}}{h}) + \frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}{h}]_0^{\sqrt 2 h}
    Edit:I have added the original solution I think something is wrong there with the natural logarithm part.
    Attached Thumbnails Attached Thumbnails How to evaluate this integral by hand?-bookssolution.jpg  
    Last edited by JohnDoe; June 19th 2011 at 02:24 PM. Reason: Equation Display
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  5. #5
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    Re: How to evaluate this integral by hand?

    try using the substitution u=\left (\tau-\frac{h}{\sqrt{2}} \right )^2
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    Re: How to evaluate this integral by hand?

    That the fourth line in the attached image contains a sum is possibly indicative of integration by parts having been used.

    CB
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    Re: How to evaluate this integral by hand?

    Quote Originally Posted by JohnDoe View Post
    Yes thanks for your help but I am stuck after these step, I cannot find the integral I might be doing something wrong here is my work:
    \frac{{\sigma 2\pi }}{{4\pi {\varepsilon _0}\sqrt 2 }}\int\limits_0^{\sqrt 2 h} {\frac{{\tau d\tau }}{{\sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}} \\<br />
{\tau ^2} + {h^2} - \sqrt 2 h\tau  = {(\tau  - \frac{h}{{\sqrt 2 }})^2} + \frac{{{h^2}}}{2}\\<br />
u = (\tau  - \frac{h}{{\sqrt 2 }})\\<br />
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_a^b {\frac{{u + \frac{h}{{\sqrt 2 }}du}}{{\sqrt {{u^2} + \frac{{{h^2}}}{2}} }}} \\
    No more \pi

    \frac{\sigma }{{{\varepsilon _0}2\sqrt 2 }}\int\limits_a^b {\frac{{u + \frac{h}{{\sqrt 2 }}du}}{{\sqrt {{u^2} + \frac{{{h^2}}}{2}} }}}

    Quote Originally Posted by JohnDoe View Post
    \frac{h}{{\sqrt 2 }}\tan \theta  = u\\<br />
\frac{\sigma }{{\pi {\varepsilon _0}2\sqrt 2 }}\int\limits_c^d {\frac{{\frac{h}{{\sqrt 2 }}(1 + \tan \theta )\frac{h}{{\sqrt 2 }}{{\sec }^2}\theta d\theta }}{{\frac{h}{{\sqrt 2 }}\sec \theta }}}<br />
    \frac{\sigma }{{{\varepsilon _0}2\sqrt 2 }}\int\limits_c^d {\frac{{\frac{h}{{\sqrt 2 }}(1 + \tan \theta )\frac{h}{{\sqrt 2 }}{{\sec }^2}\theta d\theta }}{{\frac{h}{{\sqrt 2 }}\sec \theta }}}

    Quote Originally Posted by JohnDoe View Post
    \frac{\sigma }{{\pi {\varepsilon _0}4}}\int\limits_c^d {(\sec \theta  + \sec \theta \tan \theta )d\theta }  = \frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{u^2} + {h^2}} }}{h} + \frac{{\sqrt 2 u}}{h}) + \frac{{\sqrt {2{u^2} + {h^2}} }}{h}]\\<br />
    h is missing at the numerator

    \frac{\sigma h}{{{\varepsilon _0}4}}\int\limits_c^d {(\sec \theta  + \sec \theta \tan \theta )d\theta }  = \frac{\sigma h}{{{\varepsilon _0}4}} \left[\ln (\frac{{\sqrt {2{u^2} + {h^2}} }}{h} + \frac{{\sqrt 2 u}}{h}) + \frac{{\sqrt {2{u^2} + {h^2}} }}{h} \right]

    Quote Originally Posted by JohnDoe View Post
    \frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt {2{{(\tau  - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} + \frac{{\sqrt 2 (\tau  - \frac{h}{{\sqrt 2 }})}}{h}) + \frac{{\sqrt {2{{(\tau  - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h}]_0^{\sqrt 2 h}\\<br />
\frac{\sigma }{{\pi {\varepsilon _0}4}}[\ln (\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau }  + \sqrt 2 \tau  - h}}{h}) + \frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}{h}]_0^{\sqrt 2 h}
    \frac{\sigma h}{{{\varepsilon _0}4}} \left[\ln (\frac{{\sqrt {2{{(\tau  - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} + \frac{{\sqrt 2 (\tau  - \frac{h}{{\sqrt 2 }})}}{h}) + \frac{{\sqrt {2{{(\tau  - \frac{h}{{\sqrt 2 }})}^2} + {h^2}} }}{h} \right]_0^{\sqrt 2 h}\\<br />
\frac{\sigma h}{{{\varepsilon _0}4}} \left[\ln (\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau }  + \sqrt 2 \tau  - h}}{h}) + \frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau } }}{h} \right]_0^{\sqrt 2 h}

    Your calculation is correct, you just need to finish it to finally find \frac{\sigma h}{{{\varepsilon _0}2}} \ln{\left(1+\sqrt 2 \right)}
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  8. #8
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    Re: How to evaluate this integral by hand?

    Thanks for your help, but I am still unable to understand the fourth line in the picture I have attached. Because my solution expression before the evaluation of the integral does not match the solution printed on the manual. I got h in the demoninator inside ln, whereas the solution manual does not have any h in the denominator inside ln function. What may be the reason behind that? Is it possible to reach the answer using different indefinite integrals? Can someone provide the skipped steps?
    Last edited by JohnDoe; June 20th 2011 at 03:04 PM.
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  9. #9
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    Re: How to evaluate this integral by hand?

    Quote Originally Posted by JohnDoe View Post
    I got h in the demoninator inside ln, whereas the solution manual does not have any h in the denominator inside ln function.
    It does not matter because

    \left[\ln \left(\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau }  + \sqrt 2 \tau  - h}}{h}\right)\right]_0^{\sqrt 2 h} = \left[\ln \left({\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau }  + \sqrt 2 \tau  - h}\right) - \ln(h)\right]_0^{\sqrt 2 h}

    \ln(h) is a constant that cancels out

    Therefore

    \left[\ln \left(\frac{{\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau }  + \sqrt 2 \tau  - h}}{h}\right)\right]_0^{\sqrt 2 h} = \left[\ln \left({\sqrt 2 \sqrt {{\tau ^2} + {h^2} - \sqrt 2 h\tau }  + \sqrt 2 \tau  - h}\right)\right]_0^{\sqrt 2 h}
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