# Thread: Area between two curves.

1. ## Area between two curves.

Hey

i hope this is the correct stop that i left this thread. its my last question for the day.

Find the exact area enclosed between the curve y = sin x and the line y = 1/2 for the domain 0 ≤ x ≤ 2π

answer: 3√3 - π / 3

2. Originally Posted by maay
Hey

i hope this is the correct stop that i left this thread. its my last question for the day.

Find the exact area enclosed between the curve y = sin x and the line y = 1/2 for the domain 0 ≤ x ≤ 2π

answer: 3√3 - π / 3
The line y= 1/2 crosses the curve y= sin x when sin x= 1/2 which is at $x= \pi/6$, $x= \pi- \pi/6= 6\pi/6$, $\pi+ \pi/6= 7\pi/6$ and $2\pi- \pi/6= 11\pi/6$. Also, perhaps by looking at a graph, you should be able to see that some of those areas are exactly the same and that the total area is 4 times the area from x= 0 to $\pi/6$ plus 2 times the area from $\pi/6$ to $\5\pi/6$. Those two have to be done separately because between x= 0 and $\pi/6$ 1/2 is lower than sin x and between $x= \pi/6$ and $5\pi/6$, 1/2 is higher than sin x.

Now, since you put this in "geometry" rather than "calculus", what do you know how to do? If you have learned how to "integrate" then you should know this area is $4\int_0^{\pi/6} sin(x)- 1/2 dx+ 2\int_{\pi/6}^{5\pi/6} 1/2- sin(x) dx$.

If you do not know how to integrate then I don't think it will be possible for you to do this problem.

3. Originally Posted by HallsofIvy
The line y= 1/2 crosses the curve y= sin x when sin x= 1/2 which is at $x= \pi/6$, $x= \pi- \pi/6= 6\pi/6$, $\pi+ \pi/6= 7\pi/6$ and $2\pi- \pi/6= 11\pi/6$. Also, perhaps by looking at a graph, you should be able to see that some of those areas are exactly the same and that the total area is 4 times the area from x= 0 to $\pi/6$ plus 2 times the area from $\pi/6$ to $\5\pi/6$. Those two have to be done separately because between x= 0 and $\pi/6$ 1/2 is lower than sin x and between $x= \pi/6$ and $5\pi/6$, 1/2 is higher than sin x.

Now, since you put this in "geometry" rather than "calculus", what do you know how to do? If you have learned how to "integrate" then you should know this area is $4\int_0^{\pi/6} sin(x)- 1/2 dx+ 2\int_{\pi/6}^{5\pi/6} 1/2- sin(x) dx$.

If you do not know how to integrate then I don't think it will be possible for you to do this problem.

hey yeh i know how to integrate, there is no integration thread area where you can post.

4. ## Re: Area

Originally Posted by maay
hey yeh i know how to integrate, there is no integration thread area where you can post.
Questions requiring integration belong in the Calculus subforum, which is where this thread has been moved.