# find the area bounded by the curves.

• May 12th 2011, 07:43 AM
lollycc
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
• May 12th 2011, 08:04 AM
Sudharaka
Quote:

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 $x_0=\frac{\pi}{3}$
• May 12th 2011, 08:09 AM
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.
• May 12th 2011, 08:17 AM
Sudharaka
Quote:

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.(Headbang) Thanks for pointing out the mistake.
• May 12th 2011, 08:25 AM
Ackbeet
Quote:

Originally Posted by Sudharaka
Dear Ackbeet,

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

Sure, no problem. Happens to the best of us.
• May 13th 2011, 04:36 AM
lollycc
Quote:

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 (Bow)
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 (Wondering)
• May 13th 2011, 05:00 AM
Ackbeet
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?
• May 13th 2011, 05:29 AM
lollycc
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 ...(Giggle)~ now i Can totally work this out thx so much (Clapping)
• May 13th 2011, 05:35 AM
Ackbeet
You're welcome. Let me know if you have any further difficulties.
• May 14th 2011, 09:25 PM
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 :
(Wondering) I think region 2 is right but i am not sure whether should i include the region 1 or not
• May 16th 2011, 12:33 AM
Ackbeet
Quote:

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: