Results 1 to 13 of 13

Math Help - Find the area of the region bounded between paraboa and line

  1. #1
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Find the area of the region bounded between paraboa and line

    Question : Find the area between the parabola y^2 = 4ax and the line y = mx

    Solution :
    I tried to find the intersection of the two but dont know how the equation i am getting is
    (mx)^2 = 4ax
    (mx)^2 - 4ax = 0

    I am stuck here!!!!!!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by zorro View Post
    Question : Find the area between the parabola y^2 = 4ax and the line y = mx

    Solution :
    I tried to find the intersection of the two but dont know how the equation i am getting is
    (mx)^2 = 4ax
    (mx)^2 - 4ax = 0

    I am stuck here!!!!!!!!!
    You're on the right track.

    m^2 x^2 -4ax=0

    x(m^2 x-4a)=0

    Now you can solve for the intersection points, which are you integral bounds.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Is this right

    Quote Originally Posted by Jameson View Post
    You're on the right track.

    m^2 x^2 -4ax=0

    x(m^2 x-4a)=0

    Now you can solve for the intersection points, which are you integral bounds.


    From the looks of it .It seems the intersection point are x = 0 and x = \frac{4a}{m^2}

    Am i correct!!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by zorro View Post
    From the looks of it .It seems the intersection point are x = 0 and x = \frac{4a}{m^2}

    Am i correct!!!!
    Looks good so far. Now you must integrate.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    I am stuck here as well

    Quote Originally Posted by Jameson View Post
    Looks good so far. Now you must integrate.

    Mate after integrating the equation i am getting
    \frac{4}{3} (4a^2)^{\frac{3}{2}} - 8 a^2

    Is this right!!

    \frac{4}{3} (4a)^3 - 8a^2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by zorro View Post
    Mate after integrating the equation i am getting
    \frac{4}{3} (4a^2)^{\frac{3}{2}} - 8 a^2

    Is this right!!

    \frac{4}{3} (4a)^3 - 8a^2
    Can you post your work? If integrating from x=a to x=b, you have to rewrite the parabola as y=\sqrt{4ax}=(2\sqrt{a})\sqrt{x}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    This is what i have done till now

    Quote Originally Posted by Jameson View Post
    Can you post your work? If integrating from x=a to x=b, you have to rewrite the parabola as y=\sqrt{4ax}=(2\sqrt{a})\sqrt{x}
    Since parabola is y^2 = 4ax and st. line is y = mx

    \int_{0}^{\frac{4a}{mx}} (2 \sqrt{ax}) - (mx)

    \left[ 2 \frac{(a \frac{4a}{mx})^{\frac{3}{2}}}{\frac{3}{2}} - 0 \right] - \left[ \frac{(\frac{4a}{x})^2}{2} - 0 \right]

    Is this right!!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    \int_{0}^{\frac{4a}{m^2}} [(2 \sqrt{ax}) - (mx)] dx= \left[ 2 \sqrt{a} \frac{( \frac{4a}{m^2})^{\frac{3}{2}}}{\frac{3}{2}} - 0 \right] - m \left[ \frac{(\frac{4a}{m^2})^2}{2} - 0 \right]
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Is this the end result mate

    Quote Originally Posted by dedust View Post
    \int_{0}^{\frac{4a}{m^2}} [(2 \sqrt{ax}) - (mx)] dx= \left[ 2 \sqrt{a} \frac{( \frac{4a}{m^2})^{\frac{3}{2}}}{\frac{3}{2}} - 0 \right] - m \left[ \frac{(\frac{4a}{m^2})^2}{2} - 0 \right]

    After ur previous reply i have tried to formulate the equation and this is what i am getting is this correct

    <br />
\frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{4a^2}{2m^3}<br />


    what should i do now ......or is this that end result mate
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zorro View Post
    After ur previous reply i have tried to formulate the equation and this is what i am getting is this correct

    <br />
\frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{4a^2}{2m^3}<br />


    what should i do now ......or is this that end result mate
    I haven't checked the earlier details but I can tell you that the second term is wrong. If you post your working I'll point out the mistake (although I suspect you will find the mistake yourself once you start typing the working out).
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    Quote Originally Posted by zorro View Post
    After ur previous reply i have tried to formulate the equation and this is what i am getting is this correct

    <br />
\frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{4a^2}{2m^3}<br />


    what should i do now ......or is this that end result mate
    it should be
    <br />
\frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{16a^2}{2m^3}<br />
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    I haven't checked the earlier details but I can tell you that the second term is wrong. If you post your working I'll point out the mistake (although I suspect you will find the mistake yourself once you start typing the working out).

    <br />
\frac{4}{3} \sqrt{a} \left[ \frac{4a}{m^2} \right]^{\frac{3}{2}} - \frac{8a^2}{m^3}<br />


    Do i have to do anything else after this ...
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zorro View Post
    <br />
\frac{4}{3} \sqrt{a} \left[ \frac{4a}{m^2} \right]^{\frac{3}{2}} - \frac{8a^2}{m^3}<br />


    Do i have to do anything else after this ...
    Quote Originally Posted by zorro View Post
    Question : Find the area between the parabola y^2 = 4ax and the line y = mx
    Does the above question ask you to do anything else ....?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of the region bounded by the line
    Posted in the Calculus Forum
    Replies: 8
    Last Post: December 30th 2009, 06:54 AM
  2. Find the area of the region bounded
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 20th 2009, 06:49 PM
  3. Replies: 4
    Last Post: August 13th 2009, 05:27 PM
  4. find the area of the region bounded..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 11th 2007, 08:20 PM
  5. Find the area of the region bounded in polar
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 15th 2007, 07:22 PM

Search Tags


/mathhelpforum @mathhelpforum