- #1

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## Homework Statement

Okay, I don't actually have a homework problem that explicitly tells me to find the reason why; it's just something I was wondering while I was trying to integrate circles using trigonometric substitution. But just for reference's sake: "Why does A

_{circle}= ∫√(a

^{2}-x

^{2})dx?" Additionally, when I try to take the integral of the negative formula for a semicircle, it ends up with a negative area; it's as if the circle is below the x-axis.

## Homework Equations

##x^2+y^2=a^2##

##a=radius##

##y=±\sqrt{a^2-x^2}##

## The Attempt at a Solution

##f(x)=∫\sqrt{a^2-x^2}dx##

##x=asinθ##

##dx=acosθdθ##

##θ=arcsin(\frac{x}{a})##

##f(θ)=∫\sqrt{a^2-a^2sin^2θ}⋅acosθdθ=∫\sqrt{a^2cos^2θ}⋅acosθdθ=a^2∫cosθ⋅cosθdθ##

##f(θ)=\frac{1}{2}a^2∫(1+cos2θ)dθ=\frac{1}{2}a^2(θ+\frac{1}{2}sin2θ)##

##f(θ)=\frac{1}{2}a^2(θ+sinθ⋅cosθ)##

##f(x)=\frac{1}{2}a^2(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))##

Now how do I equate...

##πa^2=\frac{1}{2}a^2(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))##

If I do some manipulation, I apparently get the the circumference of a circle.

##2πa=a(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))##

I tried doing this with real numbers earlier, which is why I'm asking this question.

##f(x)=∫\sqrt{16-x^2}dx##

##x=4sinθ##

##dx=4cosθdθ##

##θ=arcsin(\frac{x}{4})##

##f(θ)=16∫cos^2θdθ=8∫(1+cos2θ)dθ=8θ+8sinθ⋅cosθ##

##f(x)=(8arcsin(\frac{x}{4})+\frac{1}{2}x\sqrt{16-x^2})## on the interval (-4,4)

##f(4)=4π##

##f(-4)=12π##

##f(4)+f(-4)=16π##

Which is clearly the area of a circle with a radius of 4. So I integrated a semicircle and got the area of a circle. What gives?