Find the area of the region bounded between paraboa and line

**zorro** · December 1st 2009, 03:45 PM

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!!!!!!!!!

**Jameson** · December 1st 2009, 04:23 PM

Originally Posted by zorro

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.

**zorro** · December 1st 2009, 05:19 PM

Originally Posted by Jameson

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!!!!

**Jameson** · December 1st 2009, 05:22 PM

Originally Posted by zorro

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.

**zorro** · December 1st 2009, 05:59 PM

Originally Posted by Jameson

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!!

$\frac{4}{3} (4a)^3 - 8a^2$

**Jameson** · December 1st 2009, 06:41 PM

Originally Posted by zorro

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 $y=\sqrt{4ax}=(2\sqrt{a})\sqrt{x}$

**zorro** · December 1st 2009, 07:27 PM

Originally Posted by Jameson

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!!!!

**dedust** · December 1st 2009, 10:17 PM

$\int_{0}^{\frac{4a}{m^2}} [(2 \sqrt{ax}) - (mx)] dx$ = $\left[ 2 \sqrt{a} \frac{( \frac{4a}{m^2})^{\frac{3}{2}}}{\frac{3}{2}} - 0 \right] - m \left[ \frac{(\frac{4a}{m^2})^2}{2} - 0 \right]$

**zorro** · December 1st 2009, 11:21 PM

Originally Posted by dedust

$\int_{0}^{\frac{4a}{m^2}} [(2 \sqrt{ax}) - (mx)] dx$ = $\left[ 2 \sqrt{a} \frac{( \frac{4a}{m^2})^{\frac{3}{2}}}{\frac{3}{2}} - 0 \right] - m \left[ \frac{(\frac{4a}{m^2})^2}{2} - 0 \right]$

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

$ \frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{4a^2}{2m^3} $

what should i do now ......or is this that end result mate

**mr fantastic** · December 2nd 2009, 12:57 AM

Originally Posted by zorro

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

$ \frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{4a^2}{2m^3} $

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).

**dedust** · December 2nd 2009, 03:07 AM

Originally Posted by zorro

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

$ \frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{4a^2}{2m^3} $

what should i do now ......or is this that end result mate

it should be
$ \frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{16a^2}{2m^3} $

**zorro** · December 2nd 2009, 06:27 PM

Originally Posted by mr fantastic

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).

$ \frac{4}{3} \sqrt{a} \left[ \frac{4a}{m^2} \right]^{\frac{3}{2}} - \frac{8a^2}{m^3} $

Do i have to do anything else after this ...

**mr fantastic** · December 2nd 2009, 07:17 PM

Originally Posted by zorro

$ \frac{4}{3} \sqrt{a} \left[ \frac{4a}{m^2} \right]^{\frac{3}{2}} - \frac{8a^2}{m^3} $

Do i have to do anything else after this ...

Originally Posted by zorro

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 ....?

Thread: Find the area of the region bounded between paraboa and line

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Find the area of the region bounded between paraboa and line

Is this right

I am stuck here as well

This is what i have done till now

Is this the end result mate

Is this correct?

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