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Math Help - Integrals and Centroids

  1. #1
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    Integrals and Centroids

    Heh, I got more integrals I need help with, and the still unanswered centroid question, which I'll just include here.

    Centroids:
    Find the center of mass of a semi-circular plate of radius r
    Find the volume when the plate (above) is rotated around a line along its straight side

    Integrals:

     \int \frac{\ln(x)}{\sqrt{x}}dx

    \int \frac{e^{2t}}{1+e^{4t}}dt

    \int \frac{dx}{\sqrt{x^2 - 16}}
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  2. #2
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    Quote Originally Posted by Debelius View Post
    Integrals:

     \int \frac{\ln(x)}{\sqrt{x}}dx

    \int \frac{e^{2t}}{1+e^{4t}}dt

    \int \frac{dx}{\sqrt{x^2 - 16}}

     \int \frac{\ln(x)}{\sqrt{x}}dx

    Use integration by parts:

     \int \frac{\ln(x)}{\sqrt{x}}\, dx = \ln x (2\sqrt{x}) - \int \frac{2\sqrt{x}}{x}\, dx = 2\sqrt{x} \ln x - 2\int \frac{1}{\sqrt{x}}\, dx = 2\sqrt{x} \ln x - 4\sqrt{x}+C

    -----------------------------------------------------------------------------------------------------------------------------

    \int \frac{e^{2t}}{1+e^{4t}}dt

    Use substitution u = e^{2t},

    \int \frac{e^{2t}}{1+e^{4t}}dt = \frac12\int \frac{1}{1+u^2}du = \frac12\tan^{-1} u + C = \frac12\tan^{-1} e^{2t} + C

    -----------------------------------------------------------------------------------------------------------------------------

    \int \frac{dx}{\sqrt{x^2 - 16}}

    Use substitution x = 4\sec u \Rightarrow dx = 4\sec u\tan u du

    \int \frac{dx}{\sqrt{x^2 - 16}} = \int \frac{4\sec u\tan u}{4\tan u}\, du =\int \sec u \, du = \ln |\sec u + \tan u| + C = \ln \left|\dfrac{x}{4} + \frac{\sqrt{x^2 - 16}}{4}\right| + C
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Debelius View Post
    Heh, I got more integrals I need help with, and the still unanswered centroid question, which I'll just include here.

    Centroids:
    Find the center of mass of a semi-circular plate of radius r

    Chris says: To find the center of gravity [mass], we need to know its density function \delta(x,y). If you're looking for the centroid, just use the formulas \bar x = \frac{{\iint\limits_R {x\,dA}}}<br />
{{\iint\limits_R {dA}}} and \bar y = \frac{{\iint\limits_R {y\,dA}}}<br />
{{\iint\limits_R {dA}}}  <br />


    Find the volume when the plate (above) is rotated around a line along its straight side

    Chris says: We can use the Theorem of Pappus to find the volume. First find the area of the region we're dealing with, and then multiply it by the distance traveled by the centroid [in this case 2\pi\bar y]
    Hope that this makes sense!

    --Chris

    EDIT : It should 2\pi {\color{red}\bar y}, NOT 2\pi {\color{red} \bar x} as the distance traveled by the centroid (in the second question)
    Last edited by Chris L T521; July 13th 2008 at 09:11 PM. Reason: Mistake...>_>
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Debelius View Post
    Heh, I got more integrals I need help with, and the still unanswered centroid question, which I'll just include here.

    Centroids:
    Find the center of mass of a semi-circular plate of radius r
    Find the volume when the plate (above) is rotated around a line along its straight side

    Integrals:

     \int \frac{\ln(x)}{\sqrt{x}}dx

    \int \frac{e^{2t}}{1+e^{4t}}dt

    \int \frac{dx}{\sqrt{x^2 - 16}}
    Ok for the first one here is what I would do

    \int\frac{\ln(x)}{\sqrt{x}}~dx=2\int\frac{\frac{1}  {2}\ln(x)}{\sqrt{x}} =2\int\frac{\ln\left(\sqrt{x}\right)}{\sqrt{x}}~ds  =4\int\frac{\ln\left(\sqrt{x}\right)}{2\sqrt{x}}=4  \bigg[\sqrt{x}\ln(\sqrt{x})-\sqrt{x}\bigg]

    That last part was by letting u=\sqrt{x}
    ------------------------------------------------------------------------

    For the second one rewrite it as

    \int\frac{e^{2x}}{1+\left(e^{2x}\right)^2}

    Now let u=e^{2x}

    So du=2e^{2x}

    So we have

    \frac{1}{2}\int\frac{du}{1+u^2}=\frac{1}{2}\arctan  (u)

    Now just backsub
    ------------------------------------------------------------------------


    For the last one Let x=4\sec(\theta)


    I chose this because

    \left(4\sec(\theta)\right)^2-16=16\sec^2(\theta)-16=16\left(\sec^2(\theta)-1\right)=16\tan^2(\theta)
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
     \int \frac{\ln(x)}{\sqrt{x}}dx

    Use integration by parts:

     \int \frac{\ln(x)}{\sqrt{x}}\, dx = \ln x (2\sqrt{x}) - \int \frac{2\sqrt{x}}{x}\, dx = 2\sqrt{x} \ln x - 2\int \frac{1}{\sqrt{x}}\, dx = 2\sqrt{x} \ln x - 4\sqrt{x}+C
    Hmm...I didn't even think to use integration by parts
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    Hope that this makes sense!

    --Chris
    That's the problem, I don't understand how to even do those integrals for the first part. I have those equations, but I was lost.

    The second part makes sense, so if I had the equation for a half circle, I'd just do 2\pi \int f(x)dx f(x) being the equation for a circle?
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Debelius View Post
    That's the problem, I don't understand how to even do those integrals for the first part. I have those equations, but I was lost.

    The second part makes sense, so if I had the equation for a half circle, I'd just do 2\pi \int f(x)dx f(x) being the equation for a circle?
    Not quite. For the second part, you would need to do 2\pi\bar y {\iint\limits_R {dA}}

    Let me simplify the first part...

    We can rewrite the equation \bar x = \frac{{\iint\limits_R {x\,dA}}}{{\iint\limits_R {dA}}} as \bar x = \frac{1}<br />
{\text{Area of R}}{\iint\limits_R {x\,dA}} and \bar y = \frac{{\iint\limits_R {y\,dA}}}{{\iint\limits_R {dA}}} as \bar y = \frac{1}<br />
{\text{Area of R}}{\iint\limits_R {y\,dA}}

    We can determine \bar x two ways: by inspection or by evaluating the integral.

    Since the semi-circle is symmetric about the y axis, we can determine that \bar x=0. Now if you didn't see that, then we would evaluate the equation for \bar x:

    [Let me denote the radius of the circle as r_0 instead of r]...

    \bar x = \frac{1}<br />
{\text{Area of R}}{\iint\limits_R {x\,dA}}=\frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr  \,d\theta

    Note how we are using polar coordinates to evaluate the integral.

    We now see that:

    \begin{aligned}\bar x &= \frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr  \,d\theta \\<br />
 &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l  eft[\frac{1}{3}r^3\cos\theta\right]_{r=0}^{r_0} \\<br />
&= \frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}{r  _0}^3\right)\int_0^{\pi}\cos\theta \,d\theta \\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}  {r_0}^3\right)\cdot 0 \\<br />
&=\color{red}\boxed{0}\end{aligned}

    Are you still with me?


    Now let's find \bar y:

    \begin{aligned}<br />
\bar y &=\frac{1}{\frac{1}{2}\pi{r_0}^2}{\iint\limits_  R {y\,dA}}\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\int_  0^{r_0}(r\sin\theta)r\,dr\,d\theta\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l  eft[\frac{1}{3}r^3\sin\theta\right]_{r=0}^{r_0}\,d\theta\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}  {r_0}^3\right)\int_0^{\pi}\sin\theta\,d\theta\end{  aligned}
    \begin{aligned} \bar y&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3  }{r_0}^3\right)\bigg[-\cos\theta\bigg]_0^{\pi}\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{2}{3}  {r_0}^3\right)\\<br />
&=\color{red}\boxed{\frac{4r_0}{3\pi}}\end{aligned  }

    Thus, the centroid is \color{red}\boxed{\bigg(0,\frac{4r_0}{3\pi}\bigg)}

    This answers your first question.

    Now the second:

    Using the theorem of Pappus, we can find this volume:

    V=2\pi \cdot \bar y \cdot \left(\text{Area of R}\right)

    \therefore V=2\pi\cdot \left[\frac{4r_0}{3\pi}\right]\cdot \left[\frac{1}{2}\pi {r_0}^2\right]=\color{red}\boxed{\frac{4}{3}\pi{r_0}^3}

    We know this to be true because the volume of a sphere is \frac{4}{3}\pi r^3

    I hope you can absorb all of this. If you have questions, feel free to ask!

    --Chris
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Not quite. For the second part, you would need to do 2\pi\bar y {\iint\limits_R {dA}}

    Let me simplify the first part...

    We can rewrite the equation \bar x = \frac{{\iint\limits_R {x\,dA}}}{{\iint\limits_R {dA}}} as \bar x = \frac{1}<br />
{\text{Area of R}}{\iint\limits_R {x\,dA}} and \bar y = \frac{{\iint\limits_R {y\,dA}}}{{\iint\limits_R {dA}}} as \bar y = \frac{1}<br />
{\text{Area of R}}{\iint\limits_R {y\,dA}}

    We can determine \bar x two ways: by inspection or by evaluating the integral.

    Since the semi-circle is symmetric about the y axis, we can determine that \bar x=0. Now if you didn't see that, then we would evaluate the equation for \bar x:

    [Let me denote the radius of the circle as r_0 instead of r]...

    \bar x = \frac{1}<br />
{\text{Area of R}}{\iint\limits_R {x\,dA}}=\frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr  \,d\theta

    Note how we are using polar coordinates to evaluate the integral.

    We now see that:

    \begin{aligned}\bar x &= \frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr  \,d\theta \\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l  eft[\frac{1}{3}r^3\cos\theta\right]_{r=0}^{r_0} \\<br />
&= \frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}{r  _0}^3\right)\int_0^{\pi}\cos\theta \,d\theta \\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}  {r_0}^3\right)\cdot 0 \\<br />
&=\color{red}\boxed{0}\end{aligned}

    Are you still with me?


    Now let's find \bar y:

    \begin{aligned}<br />
\bar y &=\frac{1}{\frac{1}{2}\pi{r_0}^2}{\iint\limits_  R {y\,dA}}\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\int_  0^{r_0}(r\sin\theta)r\,dr\,d\theta\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l  eft[\frac{1}{3}r^3\sin\theta\right]_{r=0}^{r_0}\,d\theta\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}  {r_0}^3\right)\int_0^{\pi}\sin\theta\,d\theta\end{  aligned}
    \begin{aligned} \bar y&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3  }{r_0}^3\right)\left[-\cos\theta\right]_0^{\pi}\\<br />
&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{2}{3}  {r_0}^3\right)\\<br />
&=\color{red}\boxed{\frac{4r_0}{3\pi}}\end{aligned  }

    Thus, the centroid is \color{red}\boxed{\bigg(0,\frac{4r_0}{3\pi}\bigg)}

    This answers your first question.

    Now the second:

    Using the theorem of Pappus, we can find this volume:

    V=2\pi \cdot \bar y \cdot \left(\text{Area of R}\right)

    \therefore V=2\pi\cdot \left[\frac{4r_0}{3\pi}\right]\cdot \left[\frac{1}{2}\pi {r_0}^2\right]=\color{red}\boxed{\frac{4}{3}\pi{r_0}^3}

    We know this to be true because the volume of a sphere is \frac{4}{3}\pi r^3

    I hope you can absorb all of this. If you have questions, feel free to ask!

    --Chris
    Hmm, great work as usual Chris, but uhm talk about quantam leap from integratin \int\sin^2(\theta)\cos^3(\theta)d\theta to evaluating double integrals that need to be changed to polar coordinates to be able to be integrated by elementary terms.
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