(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the area of: {(x, y) R^2 : 0 ≤ x ≤ π, 0 ≤ y ≤ 1/2, 0 ≤ y ≤ sin x}

3. The attempt at a solution

y = sinx on the interval 0 to π but with y < 1/2.

if Sinx = 1/2, then x = pi/6, or x = 5pi/6

Because of symmetry I integrate one part and the multiply by two.

[0,pi/6) In: (sinx) dx = (-cosx) = -cos(π/6) - -cos(0) = cos(0) - cos(π/6) = 1 - (1/2)√3.

2(1 - (1/2)√3.)

I stop there, but then I saw that to the answer that I had, I must add the rectangular area from pi/6 to 1/2(pi) with height 1/2, or (1/3)pi(1/2) = pi/6

So the answer was: 2(1 - (1/2)√3 + pi/6) = 2 - √3 + (1/3)pi).

I don't understand that last part. Which rectangular area? I don't see it.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Area under de sine curve

**Physics Forums | Science Articles, Homework Help, Discussion**