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

1. ## Find the area of the region bounded between paraboa and line

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

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

I am stuck here!!!!!!!!!

2. Originally Posted by zorro
Question : Find the area between the parabola $\displaystyle y^2 = 4ax$ and the line $\displaystyle y = mx$

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

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

$\displaystyle m^2 x^2 -4ax=0$

$\displaystyle x(m^2 x-4a)=0$

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

3. ## Is this right

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

$\displaystyle m^2 x^2 -4ax=0$

$\displaystyle x(m^2 x-4a)=0$

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

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

Am i correct!!!!

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

Am i correct!!!!
Looks good so far. Now you must integrate.

5. ## I am stuck here as well

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

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

Is this right!!

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

6. Originally Posted by zorro
Mate after integrating the equation i am getting
$\displaystyle \frac{4}{3} (4a^2)^{\frac{3}{2}} - 8 a^2$

Is this right!!

$\displaystyle \frac{4}{3} (4a)^3 - 8a^2$
Can you post your work? If integrating from x=a to x=b, you have to rewrite the parabola as $\displaystyle y=\sqrt{4ax}=(2\sqrt{a})\sqrt{x}$

7. ## This is what i have done till now

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

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

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

Is this right!!!!

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

9. ## Is this the end result mate

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

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

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

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

10. 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

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

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).

11. 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

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

what should i do now ......or is this that end result mate
it should be
$\displaystyle \frac{4}{3} \sqrt{a} \left( \frac{4a}{m^2} \right)^{\frac{3}{2}} - \frac{16a^2}{2m^3}$

12. ## Is this correct?

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).

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

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

13. Originally Posted by zorro
$\displaystyle \frac{4}{3} \sqrt{a} \left[ \frac{4a}{m^2} \right]^{\frac{3}{2}} - \frac{8a^2}{m^3}$

Do i have to do anything else after this ...
Originally Posted by zorro
Question : Find the area between the parabola $\displaystyle y^2 = 4ax$ and the line $\displaystyle y = mx$
Does the above question ask you to do anything else ....?