Integral of semicircle?

Question

integral(sqrt(1-x^2))dx=?

This is where I am stuck in a larger problem involving the width of the middle stripe in a bucket of neopolitan ice cream.

It'll probably be tricky to explain this calculus using regular type, but I know that someone here will either manage that or find another way.

Answer 1

let x = cos(θ)

Answer 2

easily it does no longer. It varieties a circle... 2 infact (there is usually a particularly fainter rainbow accompanying the considerable one, which isn't many times obtrusive). the clarification we see it as a semi-circle is maximum efficient for the clarification that we (extra often than no longer) see it from the backside, so so, the backside a million/2 seems to vanish under the horizon. while you're viewing the the rainbow from a intense vantage element. Say a Tower or sky-scraper, you will see the full cirlce. i've got seen it and it blew my socks off. Rainbows are area of attractiveness. 2 little information for you: a million. No 2 persons are ever watching precisely an identical rainbow, for the clarification that in the event that they are status in 2 one-of-a-form positions (even nonetheless they are next to a minimum of one yet another) they are watching on the effect from a reasonably unique attitude. for that reason, the water droplets that are inflicting the effect for guy.a million are working otherwise for guy. 2. 2. in case you finally end up watching a rainbow, face it head on. The photograph voltaic is continuously in the present day in the back of you. without fail... If this weren't the case you does no longer be watching a rainbow... The photograph voltaic's mushy passes you and outcomes the water droplets immediately in front of you... you will continually be in between the solar and the rainbow. extremely rainbow's do no longer rather exist. they are merely a trick of the mild, that's why you will in no way in looking that pot of gold.

Answer 3

This is why it's sometimes easier switching to polar coordinates. In the polar coordinate system, as you may already know, you have r and θ such that sin(θ)=y/r, cos(θ)=x/r, r² = x² + y², etc. For the half of a unit circle like this (r=1), we can just say x = cos(θ). So dx = -sin(θ)dθ. Substitute these in to the integral, and you get:

∫√(1-cos²(θ)) (-sin(θ))dθ
∫√(sin²(θ)) (-sin(θ))dθ
∫ sin(θ)(-sin(θ))dθ
- ∫ sin²(θ)dθ

From here, you could use the identity cos(2x) = 1 - 2sin²(x) to get:

- ∫ (1/2)(cos(2θ)-1)dθ
-(1/2) ∫ (cos(2θ)-1)dθ
-(1/4) ∫ 2cos(2θ)-1 dθ
-(1/2) [ sin(2θ) - θ ]

Now substitute back in to get it in terms of x again:
-(1/2) [ 2sin(θ)cos(θ) - θ ]
-(1/2) [ 2√[1-cos²(θ)]cos(θ) - θ ]
-(1/2) [ 2√[1-x²] x - cos^-1(x) ]
-(1/2) [ 2x√[1-x²] - cos^-1(x) ]
(cos^-1(x))/2 - x√[1-x²]

So taking the semi circle (x=-1 to 1)
(-1/2)[ (-2√0 - pi) - (2√0 - 0)] =
π / 2

This is what we'd expect, because it's the area of half of a unit circle.

Answer 4

Exactly what I was going to say.

And if you have any trouble at all, just do it for positive x and double the result.