Comments on: Find the value of the integral by using an area formula from geometry? http://www.thearealist.com/area-formula/find-the-value-of-the-integral-by-using-an-area-formula-from-geometry Fri, 03 Feb 2012 13:07:59 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: Tabula Rasa http://www.thearealist.com/area-formula/find-the-value-of-the-integral-by-using-an-area-formula-from-geometry/comment-page-1#comment-1707 Tabula Rasa Fri, 22 Jan 2010 10:27:59 +0000 http://www.thearealist.com/area-formula/find-the-value-of-the-integral-by-using-an-area-formula-from-geometry#comment-1707 You should use formula for calculating area of circular segment http://en.wikipedia.org/wiki/Circular_segment 1) Integral gives area equal to half area of the circular segment for x from 1/2 to 1 of the unit circle. A = (1/2) ·(R²/2)·(θ - sin(θ)) R = 1, θ = 120° = 2π/3 A = (1/4) (2π/3 – sin(2π/3)) = (2π/3 – √3/2)/4 ≈ 0.307 If you have doubts just give me a message. Good luck!<br><b>References : </b><br>http://en.wikipedia.org/wiki/Circular_segment You should use formula for calculating area of circular segment

http://en.wikipedia.org/wiki/Circular_segment

1) Integral gives area equal to half area of the circular segment for x from 1/2 to 1 of the unit circle.

A = (1/2) ·(R²/2)·(θ – sin(θ))

R = 1, θ = 120° = 2π/3

A = (1/4) (2π/3 – sin(2π/3)) = (2π/3 – √3/2)/4 ≈ 0.307

If you have doubts just give me a message. Good luck!
References :
http://en.wikipedia.org/wiki/Circular_segment

]]> By: gintable http://www.thearealist.com/area-formula/find-the-value-of-the-integral-by-using-an-area-formula-from-geometry/comment-page-1#comment-1706 gintable Fri, 22 Jan 2010 10:12:59 +0000 http://www.thearealist.com/area-formula/find-the-value-of-the-integral-by-using-an-area-formula-from-geometry#comment-1706 Your teacher is being unnecessarily cruel by making you use the formula for the area of a circle to evaluate the integral. If you could argue that these are pie slices, it would be much easier, but they aren't. Look, the indefinate integral relative to x of sqrt(r^2 - x^2) is... ∫sqrt(r^2 - x^2) dx = 1/2*x^2*sqrt(r^2 - x^2) + 1/2*r^2*arctan(x/sqrt(r^2-x^2)) + C Call this I(x) = 1/2*x^2*sqrt(r^2 - x^2) + 1/2*r^2*arctan(x/sqrt(r^2-x^2)) + C Your arctangent function will return in radians, so have your mode set accordingly. For part 1: r = 1. Plug in x=1 and x=1/2 to the indefinite integral result. Subtract I(1) - I(1/2) and you got your result. We do need to replace the arctangent function with Pi/2, because it is the arctangent of an expression containing division by zero. 0.8307 units^2 is the answer. The analytical answer is a mess, so no need to show. For part 2: r = sqrt(2). Plug in x=1 and x=0 to the indefinite integral result. Subtract I(1) - I(0) and you got your result. 1/2 + Pi/4 gives you the answer, or 1.285 units^2.<br><b>References : </b><br> Your teacher is being unnecessarily cruel by making you use the formula for the area of a circle to evaluate the integral. If you could argue that these are pie slices, it would be much easier, but they aren’t.

Look, the indefinate integral relative to x of sqrt(r^2 – x^2) is…
∫sqrt(r^2 – x^2) dx = 1/2*x^2*sqrt(r^2 – x^2) + 1/2*r^2*arctan(x/sqrt(r^2-x^2)) + C

Call this I(x) = 1/2*x^2*sqrt(r^2 – x^2) + 1/2*r^2*arctan(x/sqrt(r^2-x^2)) + C

Your arctangent function will return in radians, so have your mode set accordingly.

For part 1: r = 1. Plug in x=1 and x=1/2 to the indefinite integral result. Subtract I(1) – I(1/2) and you got your result. We do need to replace the arctangent function with Pi/2, because it is the arctangent of an expression containing division by zero. 0.8307 units^2 is the answer. The analytical answer is a mess, so no need to show.

For part 2: r = sqrt(2). Plug in x=1 and x=0 to the indefinite integral result. Subtract I(1) – I(0) and you got your result. 1/2 + Pi/4 gives you the answer, or 1.285 units^2.
References :

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