# Math Help - find the area bounded by the curves.

1. ## find the area bounded by the curves.

find the area bounded by the curve y=cos x and the line y=0 and y=\frac{3}{ 2\pi } x could u plz show me more detail solution of this ~~~ thx

2. Originally Posted by lollycc
find the area bounded by the curve y=cos x and the line y=0 and y=\frac{3}{ 2\pi } x could u plz show me more detail solution of this ~~~ thx

Dear lollycc,

Find the intersection point of the two curves. Suppose the intersection point is $x_0$. Then you have to get the integration,

$Area=\int_{0}^{x_0}\cos x~ dx-\int_{0}^{x_0}\frac{3x}{2\pi}~ dx$

Note that, $x=\frac{\pi}{3}$ satisfies $\cos x=\frac{3x}{2\pi}$. Hence $x_0=\frac{\pi}{3}$

3. I'm not sure it's quite as simple as that, Sudharaka. Here's a plot of the two functions. It seems to me that you have to integrate the straight line up to the point of intersection, and then cosine from the intersection to pi/2.

4. Originally Posted by Ackbeet
I'm not sure it's quite as simple as that, Sudharaka. Here's a plot of the two functions. It seems to me that you have to integrate the straight line up to the point of intersection, and then cosine from the intersection to pi/2.
Dear Ackbeet,

I have mistakenly taken the y axis(x=0 instead of y=0) as a boundary curve. Thanks for pointing out the mistake.

5. Originally Posted by Sudharaka
Dear Ackbeet,

I have mistakenly taken the y axis(x=0 instead of y=0) as a boundary curve. Thanks for pointing out the mistake.
Sure, no problem. Happens to the best of us.

6. Originally Posted by Ackbeet
I'm not sure it's quite as simple as that, Sudharaka. Here's a plot of the two functions. It seems to me that you have to integrate the straight line up to the point of intersection, and then cosine from the intersection to pi/2.
thx guys for ur help ...i am really appreciated ~
Dear Ackbeet,
can u explain to me a little bit more of this question...maybe can u show me some detailed working out i am still kinda confused
i just to do the integration of cos x - integration of 3/2pi x
both from zero to their intersection ...not sure if it is right~ but i found their intersection can not be expressed in fraction it looks like kind of irrational number

7. Sudharaka found the point of intersection in Post # 2. It is pi/3. So integrate the straight line from zero to the point of intersection, and then ADD the integral of cosine from the point of intersection to pi/2, where the cosine function hits the x-axis (equivalent to y = 0, which is one of the boundaries of your region). Does that make sense?

8. great~ i c now whoopse i realised i just made the same mistake as Sudharaka did...my frd confused me by telling me the area we are looking for it bounded by the two equations and the y-axis ...~ now i Can totally work this out thx so much

9. You're welcome. Let me know if you have any further difficulties.

10. Dear Ackbeet,
I think I need ur help now~
some one said there are actually two enclosed areas, by these 3 equations. One is from -pi/2 to pi/3 , and the other one is from 0 to pi/2 . I double checked the equation it said find the AREA bounded by the curves ,so it means there just one area right? I am confused now...
The area of region 1 is:

The area of region 2 is :

I think region 2 is right but i am not sure whether should i include the region 1 or not

11. Originally Posted by lollycc
Dear Ackbeet,
I think I need ur help now~
some one said there are actually two enclosed areas, by these 3 equations. One is from -pi/2 to pi/3 , and the other one is from 0 to pi/2 . I double checked the equation it said find the AREA bounded by the curves ,so it means there just one area right? I am confused now...
The area of region 1 is:

The area of region 2 is :

I think region 2 is right but i am not sure whether should i include the region 1 or not
You should not include your "Region 1", since it's not actually a bounded, enclosed area. Referring back to the plot I linked to in Post # 3, there's only one intersection of the straight line with the cosine function. Hence, there can only be only be one enclosed area at most. Your Region 2 is correct.

12. huray ~ finally i solved this question thx for confirmed my work ...
i am appreciated for ur detailed explanation

13. You're welcome!