How can I use Reimann integral to find the area of a circle?

[Zula110100100]

Say we have a circle of unit radius centered on the origin, and another point on the x-axis d units away, were we to draw a line at angle θ relative to the x axis, cutting across the circle at either 2, 1 or 0 points, the length of the chord made(or not made) is

L(\theta) = 2r\sqrt{1-\frac{d^2\sin^2(\theta)}{r^2}}

My calculus isn't up to par to figure out how to then determine the area of the circle using that equation...I would have thought maybe using the area of sector .5r^2dθ with r = L(θ) but I keep getting the wrong answers, is it not possible to integrate over theta here and get the area of the circle? Is the above function incorrect?

 

[tiny-tim]

Hi Zula110100100!

to then determine the area of the circle using that equation...… I would have thought maybe using the area of sector .5r²dθ with r = L(θ) but I keep getting the wrong answers

Your formulas look ok, and integrating from θ = 0 to π should work …

show us what you did ​

 

[Zula110100100]

I put it all together to get:

A = \int_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}{\frac{1}{2}(4r^2-4d^2\sin^2(\theta))}d\theta = 2r^2\theta + d^2\sin(2\theta)-2d^2\theta |_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}

A = 2\sin^{-1}(\frac{1}{1.5}) + (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-2(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) - (-2\sin^{-1}(\frac{1}{1.5}) - (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))+2(1.5)^{2}\sin^{-1}(\frac{1}{1.5})) = 4\sin^{-1}(\frac{1}{1.5}) +2(1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-4(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) \approx -.65

The domain of L(θ)
1-d^2sin^2(θ)/r^2≥0
d^2sin^2(θ)/r^2≤1
sin^2(θ) ≤ r^2/d^2
sin(θ) ≤ r/d
θ <= arcsin(r/d)

 

[tiny-tim]

Hi Zula110100100!

(just got up :zzz:)

The domain of L(θ)
1-d^2sin^2(θ)/r^2≥0
d^2sin^2(θ)/r^2≤1
sin^2(θ) ≤ r^2/d^2
sin(θ) ≤ r/d
θ <= arcsin(r/d)

but r > d, so r/d > 1, so arcsin doesn't exist!

the limits are 0 to π …

the pivot point (at d) is fixed, and the line rotates from horizontal to horizontal

I put it all together to get:

A = \int_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}{\frac{1}{2}(4r^2-4d^2\sin^2(\theta))}d\theta = 2r^2\theta + d^2\sin(2\theta)-2d^2\theta |_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}

A = 2\sin^{-1}(\frac{1}{1.5}) + (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-2(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) - (-2\sin^{-1}(\frac{1}{1.5}) - (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))+2(1.5)^{2}\sin^{-1}(\frac{1}{1.5})) = 4\sin^{-1}(\frac{1}{1.5}) +2(1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-4(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) \approx -.65

no, the integrand has a square-root

however, although i said that you could solve it that way, on second thoughts, i don't think that √(r² - d²sin²θ) has a straightforward integral

is this a question from a book, or one you've made up out of interest?​

 

[Zula110100100]

out of interests...d>r, (d,0) is a point outside the circle. So arcsin(r/d) should I think somehow be the angle, relative to the x-axis, that a tangent line to the circle passes through (d,0)

Does the root not go away from the fact that I am using L(θ) as r in .5r^2θ so .5(L(θ))^2θ and the root gets squared to find the area of each sector

 

[Zula110100100]

So here is a graph on a unit circle centered at origin, y=sqrt(1-x^2) and a line that passes through (d,0), d=1.5, the slope of that line is -tan(arcsin(r/d))=-tan(arcsin(1/1.5)), as you can see it is the correct slope, so greater than that there are no points of intersection, and less than that and greater than it's negative(consider +/- sqrt circle) there are two points of intersection, L(θ) from above should give the distance between those two points of intersection, and thus a chord from the circle. I am attemepting to integrate by adding areas of sectors that are between two such chords, though that area is not truly a sector, or even too very close to one...but, I was under the impression that as dθ approached 0 it would be close enough...more importantly though, why does the sqrt not go away when being squared?

I am not sure how to shade the sector with the prgrm I am using, but it would be between the two lines.

 

[Zula110100100]

Just realized I didn't mention d>r in the first post and I really should have, does the formula still look correct for that case?

 

[Zula110100100]

Using the program to get the graphs there i put in the forumla for L(θ) and it has an option to integrate for area and it gave me 2.23...so my question is why can't I integrate for area this way? the distance from the point to the region being measured makes it impossible to use theta to integrate?

 

[tiny-tim]

out of interests...d>r, (d,0) is a point outside the circle.

ah!

Does the root not go away from the fact that I am using L(θ) as r in .5r^2θ so .5(L(θ))^2θ and the root gets squared to find the area of each sector

ooh yes, you're right, it is squared

but, for each value of θ, you need to find the area of the dθ sector centred at the "d" point to the furthest intersection, minus the sector to the nearest intersection …

your single sector based on the distance between those two points, is much smaller