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Math Help - Integrations

  1. #1
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    Integrations

    1. The curve C has parametric equation x = sin t, y = sin 2t, 0 ≤ t ≤ pi/2.

    If the region is revolved through 2 \pi radians about the x-axis, find the volume of the solid formed.
    -----------------------
    So Volume = \pi \int y^2 dx = \pi \int y^2 \frac {dx}{dt}dt

     = \pi \int_{0}^{\frac{\pi}{2}} sin^2 2t \,cos\,t\, dt

    Then how could I integrate this?
    -----------------------

    2. Find a general solution of the following differential equation.

    e^{x+y} \frac{dy}{dx} = x(2 + e^y)
    -----------------------

    So far I did,

     \int \frac{e^y}{2+e^y} dy = \int \frac{x}{e^x} dx

     \int (1 - \frac{2}{2 + e^y})dy = \int x\, e^{-1}dx

     y - 2 \,ln|2+e^y| = -xe^{-x} - e^{-x} + C


    But this is not right ans. The right ans is ln|2+e^y| = -xe^{-x} - e^{-x} + C

    So where did I wrong?
    -----------------------
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  2. #2
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    Quote Originally Posted by geton View Post
    1. The curve C has parametric equation x = sin t, y = sin 2t, 0 ≤ t ≤ pi/2.

    If the region is revolved through 2 \pi radians about the x-axis, find the volume of the solid formed.
    -----------------------
    So Volume = \pi \int y^2 dx = \pi \int y^2 \frac {dx}{dt}dt

     = \pi \int_{0}^{\frac{\pi}{2}} sin^2 2t \,cos\,t\, dt

    Then how could I integrate this?
    -----------------------

    2. Find a general solution of the following differential equation.

    e^{x+y} \frac{dy}{dx} = x(2 + e^y)
    -----------------------

    So far I did,

     \int \frac{e^y}{2+e^y} dy = \int \frac{x}{e^x} dx

     \int (1 - \frac{2}{2 + e^y})dy = \int x\, e^{-1}dx

     y - 2 \,ln|2+e^y| = -xe^{-x} - e^{-x} + C


    But this is not right ans. The right ans is ln|2+e^y| = -xe^{-x} - e^{-x} + C

    So where did I wrong?
    -----------------------
    Q1 Expand

     sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t and use a substitution  u = cos\,t\,

    Q2

     \int \frac{e^y}{2+e^y} dy \ne y - 2 \,ln|2+e^y|

    If  u = 2+e^y then  du = e^y\, dy so  \int \frac{e^y}{2+e^y} dy = \int \frac{du}{u}

    which easily interates giving

     \int \frac{e^y}{2+e^y} dy = \,ln|2+e^y| + c
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    Q1 Expand

     sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t and use a substitution  u = cos\,t\,

    Q2

     \int \frac{e^y}{2+e^y} dy \ne y - 2 \,ln|2+e^y|

    If  u = 2+e^y then  du = e^y\, dy so  \int \frac{e^y}{2+e^y} dy = \int \frac{du}{u}

    which easily interates giving

     \int \frac{e^y}{2+e^y} dy = \,ln|2+e^y| + c

    Forgive my ignorance but how did you guys seperate the e^{x+y} term?

    Thanks
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  4. #4
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    Quote Originally Posted by tsal15 View Post
    Forgive my ignorance but how did you guys seperate the e^{x+y} term?

    Thanks
    One of the properties of exponents

    e^{x+y} = e^x \, e^y
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  5. #5
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    Quote Originally Posted by geton View Post

    2. Find a general solution of the following differential equation.

    e^{x+y} \frac{dy}{dx} = x(2 + e^y)
    As you noted \begin{aligned}e^{x+y}\frac{dy}{dx}=x(2+e^y)&\impl  ies \frac{e^y}{2+e^y}dy=\frac{x}{e^x}dx\\<br />
&\implies\int\frac{e^y}{2+e^y}dy=\int\frac{x}{e^x}  dx\\<br />
&\implies \color{red}{\ln(2+e^y)}=-\left(\frac{x}{e^x}+\frac{1}{e^x}\right)+C\\<br />
&\implies y=\ln\left(\exp\left(-\left\{\frac{x}{e^x}+\frac{1}{e^x}\right\}+C\right  )-2\right)<br />
\end{aligned}

    The problem is that if you let \varphi\mapsto 2+e^y in your integral you get d\varphi=e^y~dy so you should have \int\frac{e^y}{2+e^y}dy\stackrel{\varphi=2+e^y}{\l  ongmapsto}\int\frac{d\varphi}{\varphi}
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  6. #6
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    Quote Originally Posted by danny arrigo View Post
    Q1 Expand

     sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t and use a substitution  u = cos\,t\,

    Ok,

    Let u = cos\, t
    \frac{du}{dt} = - sin\,t

    Then?

    Could you explain how to do this?
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  7. #7
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    Quote Originally Posted by geton View Post
    Ok,

    Let u = cos\, t
    \frac{du}{dt} = - sin\,t

    Then?

    Could you explain how to do this?
    So far I thought

    \int 4 \, sin^2t \,cos^3 t \,dt
    = - 4 \int u^3 \sqrt{1-u^2}\,du

    Is it the right way? Then?
    Last edited by geton; December 25th 2008 at 10:25 PM.
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  8. #8
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    Quote Originally Posted by geton View Post
    So far I thought

    ∫4 sin^2 t cos^3 t dt
    = 4∫ (1-cos^2 t)cos^3 t dt
    = 4∫ (cos^3 t - cos^5 t) dt

    Is it the right way? Then?
    No.

    First you note that 4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t.

    Then you make the substitution {\color{red}u = \sin t} (NOT u = \cos t which was no doubt a typo by danny). Note that dt = \frac{du}{\cos t}.
    Last edited by mr fantastic; December 26th 2008 at 11:52 AM.
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    As you noted \begin{aligned}e^{x+y}\frac{dy}{dx}=x(2+e^y)&\impl  ies \frac{e^y}{2+e^y}dy=\frac{x}{e^x}dx\\<br />
&\implies\int\frac{e^y}{2+e^y}dy=\int\frac{x}{e^x}  dx\\<br />
&\implies \color{red}{\ln(2+e^y)}=-\left(\frac{x}{e^x}+\frac{1}{e^x}\right)+C\\<br />
&\implies y=\ln\left(\exp\left(-\left\{\frac{x}{e^x}+\frac{1}{e^x}\right\}+C\right  )-2\right)<br />
\end{aligned}

    The problem is that if you let \varphi\mapsto 2+e^y in your integral you get d\varphi=e^y~dy so you should have \int\frac{e^y}{2+e^y}dy\stackrel{\varphi=2+e^y}{\l  ongmapsto}\int\frac{d\varphi}{\varphi}
    My b and b.

    There's not really problem.

    The result given by danny is correct. What you've done (which is fine by the way) is to make explicit how the integration is done.
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    No.

    First you note that 4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t.

    Then you make the substitution {\color{red}u = \sin t} (NOT u = \sin t which was no doubt a typo by danny). Note that dt = \frac{du}{\cos t}.
    Thanks.
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  11. #11
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    Quote Originally Posted by mr fantastic View Post
    My b and b.

    There's not really problem.

    The result given by danny is correct. What you've done (which is fine by the way) is to make explicit how the integration is done.
    Yes. I am sorry. I missed Danny's post.
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    No.

    First you note that 4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t.

    Then you make the substitution {\color{red}u = \sin t} (NOT u = \cos t which was no doubt a typo by danny). Note that dt = \frac{du}{\cos t}.
    Yep - a typo on my part. It should be u = \sin t
    Last edited by mr fantastic; December 26th 2008 at 11:53 AM.
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